{"id":13265,"date":"2024-03-26T10:15:14","date_gmt":"2024-03-26T04:45:14","guid":{"rendered":"https:\/\/gstguntur.com\/?p=13265"},"modified":"2024-04-02T17:57:57","modified_gmt":"2024-04-02T12:27:57","slug":"basic-concepts-of-permutations-and-combinations-ca-foundation-maths-study-material","status":"publish","type":"post","link":"https:\/\/gstguntur.com\/basic-concepts-of-permutations-and-combinations-ca-foundation-maths-study-material\/","title":{"rendered":"Basic Concepts of Permutations and Combinations \u2013 CA Foundation Maths Study Material"},"content":{"rendered":"

This Basic Concepts of Permutations and Combinations \u2013 CA Foundation Maths Study Material<\/a>\u00a0is designed strictly as per the latest syllabus and exam pattern.<\/p>\n

Basic Concepts of Permutations and Combinations \u2013 CA Foundation Maths Study Material<\/h2>\n

Permutations and combinations :<\/strong>
\nThere are various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a Permu\u00actation when the order of selection is a factor and this selection of subsets is called a Combination when order is not a factor.
\nThe concepts of and differences between permutations and combinations can be illustrated by examination of all the different ways in which a pair of objects can be selected from five distinguishable objects such as the letters A, B and C. If both the letters selected and the order of selection are considered, then the following 6 outcomes are possible they are AB ; AC ; BC; BA; CA; CB<\/p>\n

Each of these 6 different possible selections is called a permutation. In particular, they are called the permutations of three objects taken two at a time, and the number of such permutations possible is denoted by the symbol3 P2, read \u201c3 permute 2.\u201d In general, if there are n objects available from which to select, and permutations (P) are to be formed using r of the objects at a time, the number of different permutations possible is denoted by the symbol nPr & formulated as
\nn<\/sup>Pr<\/sub> = \\(\\frac{n !}{(n-r) !}\\)
\n3<\/sup>P2<\/sub> = \\(\\frac{3 !}{(3-2) !}\\) = 3! = 3.2.1 = 6
\nFor Combinations, r objects are selected from a set of n objects to produce subsets without ordering. Contrasting the previous permutation example with the corre\u00acsponding combination, the AB and BA subsets are no longer distinct selections; by eliminating such cases there remain only 3 different possible subsets\u2014AB, AC, BC.<\/p>\n

The number of such subsets is denoted by n<\/sup>Cr<\/sub>, read \u201cn choose r\u201d & is formulated as n<\/sup>Cr<\/sub> = \\(\\frac{n !}{r !(n-r) !}\\)<\/p>\n

Fundamental Rule of Counting<\/strong>
\n(i) The SUM Rule\/Addition Rule
\nEx- 7 buses
\nCP Laxmi Nagar
\n5 Autoes
\nThere are 7 buses and 5 Autoes to go from CP to Laxmi Nagar. In how many ways a person can reach Laxmi Nagar from CP by buses or autoes. We have two options \/choices of vehicles Bus or Auto. There are 7 options to reach by bus or 5 options to reach by auto.
\n\u2234 Total ways to reach Laxmi Nagar from CP
\n= 7 + 5 = 12 ways.
\nRule: If there are two alternative jobs,one can be done by \u201cp\u201d ways and another by \u201cq\u201d ways, either of these two jobs can be done by (p + q) ways.<\/p>\n

(ii) Multiplication (Product) Rule:
\nIf one things is done by \u201cp\u201d different ways and when it has been done a second thing can be done by \u201cq\u201d different ways then total number of ways of doing both things simultaneously
\n= p \u00d7 q. ways<\/p>\n

Ex-1 Suppose, there are 7 different buses to reach Jaipur from Delhi. In how many ways a person can go to Jaipur by a bus and can come back by different bus. There are 7 options to go to Jaipur by bus but 6 options to come back by a different bus.
\n\u2234 Total No. of ways to reach Jaipur and come back
\n= 7 \u00d7 6 = 42 ways<\/p>\n

Ex-2 There are 5 roads to reach LUCKNOW from Delhi; 4 roads to reach Gorakhpur from Lucknow and 6 roads for Patna from Gorakhpur. In how many different ways can a person reach Patna from Delhi.
\nAnswer:Total No. of ways = 5 \u00d7 4 \u00d7 6 = 120
\nNote:
\n(i) In case of either or; or; one of them; atleast one; or options then use addition rule
\n(ii) In case of AND ; then use product rule.<\/p>\n

\"Basic<\/p>\n

Previous Year Exam Questions<\/strong><\/p>\n

Question 1.
\nThe number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the same straight line, is: [1 Mark, Nov. 2006]
\n(a) 185
\n(b) 175
\n(c) 115
\n(d) 105
\nAnswer:
\n(a) is Correct
\nNo. of triangles = 12<\/sup>C3<\/sub> – 7<\/sup>C3<\/sub>
\n= \\(\\frac{12 !}{3 ! 9 !}-\\frac{7 !}{3 ! 4 !}\\) = 220 – 35
\n= 185<\/p>\n

Question 2.
\nA code word is to consist of two English alphabets followed by two distinct numbers from 1 to 9. How many such code words are there? [1 Mark, Nov. 2006]
\n(a) 6,15,800
\n(b) 46,800
\n(c) 7,19,500
\n(d) 4,10,800
\nAnswer:
\n(b) No. of code Words = 26<\/sup>P2<\/sub> \u00d7 9<\/sup>P2<\/sub>
\n= \\(\\frac{26 \\cdot 25 \\cdot 24 !}{24 !} \\times \\frac{9 !}{7 !}\\)
\n= 650 \u00d7 72 = 46,800<\/p>\n

Question 3.
\nA boy has 3 library tickets and 8 books of his interest in the libraiy. Of these 8, he does not want to borrow Mathematics part-II unless Mathematics part -1 is also borrowed? In how many ways can he choose the three books to be borrowed? [1 Mark, Nov. 2006]
\n(a) 41
\n(b) 51
\n(c) 61
\n(d) 71
\nAnswer:
\n(a) is correct
\n\"Basic
\nTotal Combinations = 6+15 + 20 = 41<\/p>\n

Question 4.
\nAn examination paper consists of 12 questions divided into two parts A and B. Part A contains 7 questions and part B contains 5 questions. A candidate is required to attempt 8 questions selecting at least 3 from each part. In how many maximum ways can the candidate select the questions ? [1 Mark, Feb. 2007]
\n(a) 35
\n(b) 175
\n(c) 210
\n(d) 420
\nAnswer:
\n(d) is correct 12 Qts.
\n\"Basic
\nTotal No. of ways to select Qts.
\n= 7<\/sup>C3<\/sub>.5<\/sup>C5<\/sub> + 7<\/sup>C4<\/sub>.5<\/sup>C4<\/sub> + 7<\/sup>C5<\/sub>.5<\/sup>C3<\/sub>
\n= 35 \u00d7 1 + 35 \u00d7 5 + 21 \u00d7 10
\n= 35 + 175 + 210 = 420<\/p>\n

Question 5.
\nA Supreme Court Bench consists of 5 judges. In how many ways, the bench can give a majority division ? [1 Mark, Feb. 2007]
\n(a) 10
\n(b) 5
\n(c) 15
\n(d) 16
\nAnswer:
\n(d) is correct<\/p>\n

No. of ways to give majority decision
\n= 5<\/sup>C3<\/sub> + 5<\/sup>C4<\/sub> + 5<\/sup>C3<\/sub>
\n= 10 + 5 + 1
\n= 16<\/p>\n

Question 6.
\nGiven: P (7, k) = 60 P(7, k – 3). Then: [1 Mark, Feb. 2007]
\n(a) k = 9
\n(b) k = 8
\n(c) k = 5
\n(d) k = 0
\nAnswer:
\n(c) is Correct
\nP(7, K) = 60.p(7, k-3)
\n\"Basic
\nor (10 – k) (9 – k) (8 – k) = 60
\nor (10 – k) (0 – k) (8 – k) = 5.4.3
\n[make 3 factors of 60 so that their differences are 1.]
\nComparing on both sides ; we get
\n10 – k = 5
\nor k= 10 – 5 = 5
\n\u2234 k = 5<\/p>\n

\"Basic<\/p>\n

Question 7.
\nThe number of ways in which n books can be arranged on a shelf so that two particular books are not together is : [1 Mark, Feb. 2007]
\n(a) (n – 2) \u00d7 (n – 1)!
\n(b) (n – 2) \u00d7 (n + l)!
\n(c) (n – 1) \u00d7 (n + 1)!
\n(d) (n – 2) \u00d7 (n + 2)!
\nAnswer:
\n(a) is correct
\nTotal No. of ways to arrange book as Qts.
\n= n! – (n – 2 + 1)! .2 !
\n= n! – (n – 1)!.2!
\n= n (n – 1)! – 2 (n – 1)!
\n= (n – 1)!. (n – 2)
\n= (n – 2). (n – 1)!<\/p>\n

Question 8.
\nIn how many ways can the letters of the word FAILURE be arranged so that the consonants may occupy only odd positions? [1 Mark, May 2007]
\n(a) 576
\n(b) 476
\n(c) 376
\n(d) 276
\nAnswer:
\n(a) is correct Consonants = F, L, R
\nVowels = A, I, U, E
\nTotal letters = 7
\nThere will be 7 Positions of letters.
\nOdd positions are = 1, 3, 5, 7
\nEven positions are = 2, 4, 6
\nFrom Questions
\nNo. of Words = 4<\/sup>P3<\/sub> .4! = 4! \u00d7 4!
\n= 24 \u00d7 24 = 576<\/p>\n

Question 9.
\nFive bulbs of which three are defective are to be tried in two lights-points in a dark-room. In how many trials the room shall be lighted? [1 Mark, May 2007]
\n(a) 10
\n(b) 1
\n(c) 3
\n(d) None of these
\nAnswer:
\n(b) is correct No. of bulbs = 5
\nDefective bulbs = 3
\nNon-Defective bulbs = 2
\nNo. of Point in the room = 2
\nNo. of ways to light the darkroom
\n= 3<\/sup>C1<\/sub>.2<\/sup>C1<\/sub> + 3<\/sup>C0<\/sub>.2<\/sup>C2<\/sub>
\n= 3 \u00d7 2 + 1 \u00d7 1
\n= 6 + 1 = 7<\/p>\n

Question 10.
\nIn how many ways can a party of 4 men and 4 women be seated at a circular table, so that no two women are adjacent ? [1 Mark, May 2007]
\n(a) 164
\n(b) 174
\n(c) 144
\n(d) 154
\nAnswer:
\n(c) is correct.
\nNo. of ways to sit 4 men around a table = (4 – 1)! = 6
\nNo. of ways to sit 4 women = 4! = 24
\n\u2234 Total Ways to arrange = 6 \u00d7 24 = 144<\/p>\n

Question 11.
\nThe value of \\(\\sum_{r=1}^5\\) 5<\/sup>Cr<\/sub> is : [1 Mark, May 2007]
\n(a) 29
\n(b) 31
\n(c) 35
\n(d) 26
\nAnswer:
\n(b) is correct
\n\\(\\sum_{r=1}^5\\) 5<\/sup>Cr<\/sub> = 25<\/sup> -1 = 31<\/p>\n

Question 12.
\nIf 6<\/sup>Pr<\/sub> = 24 6<\/sup>Cr<\/sub>, then find r : [1 Mark, Aug. 2007]
\n(a) 4
\n(b) 6
\n(c) 2
\n(d) 1
\nAnswer:
\n(a) is correct
\n6<\/sup>Pr<\/sub> = 24.6<\/sup>Cr<\/sub>
\nor \\(\\frac{6 !}{(6-r) !}=\\) = 24.\\(\\frac{6 !}{r !(6-r) !}\\)
\nor r! = 24 or r! = 4! r = 4<\/p>\n

\"Basic<\/p>\n

Question 13.
\nFind the number of combinations of the letters of the word COLLEGE taken four together: [1 Mark, Aug. 2007]
\n(a) 18
\n(b) 16
\n(c) 20
\n(d) 26
\nAnswer:
\n(a) is correct
\nLetters are C, O, L, L, E, E, G
\nCase I 2 same & 2 same letters = LLEE = Only 1 combination
\nCase II LL and 2 out of C, O, E, G = 1 \u00d7 4<\/sup>C2<\/sub> = 6
\nEE and any 2 out of C, O, L, G = 1 \u00d7 4<\/sup>C2<\/sub> = 6
\nCase III All different letters = Any 4 out of C, O, L, E, G = 5<\/sup>C4<\/sub> = 5
\nTotal No. of combinations = 1 + (6 + 6) + 5 = 18<\/p>\n

Question 14.
\nHow many words can be formed with the letters of the word \u2018ORIENTAL\u2019 so that A and E always occupy odd places: [1 Mark, Aug. 2007]
\n(a) 540
\n(b) 8640
\n(c) 8460
\n(d) 8450
\nAnswer:
\n(b) is correct
\nAs Qts. No. of Words = 4<\/sup>P2<\/sub> \u00d7 6!
\n(Because there are 4 Positions 1, 3, 5, 7 for A & E)
\n= \\(\\frac{4 !}{(4-2) !}\\) \u00d7 720 = 12 \u00d7 720 = 8640<\/p>\n

Question 15.
\nIf 1000<\/sup>C98<\/sub> = 999<\/sup>C97<\/sub> + x<\/sup>C901<\/sub>, find x : [1 Mark, Nov. 2007]
\n(a) 999
\n(b) 998
\n(c) 997
\n(d) 1000
\nAnswer:
\n(a) is correct
\nTricks : Go by choices
\nFor (a) RHS = 999<\/sup>C97<\/sub> + 999<\/sup>C901<\/sub> (If x = 999 )
\n= 999<\/sup>C97<\/sub> + 999<\/sup>C98<\/sub> = 1000<\/sup>C98<\/sub> = LHS<\/p>\n

Question 16.
\nHow many numbers greater than a million can be formed with the digits 4, 5, 5, 0, 4, 5, 3? [Nov. 2007]
\n(a) 260
\n(b) 360
\n(c) 280
\n(d) 380
\nAnswer:
\n(b) is correct
\nNo. of Numbers = \\(\\frac{7 !}{3 ! .2 !}\\) – 1 \u00d7 \\(\\frac{6 !}{3 ! .2 !}\\)
\n[Zero is fixed at 1st positions with permutation 1]
\n= 420 – 60 = 360<\/p>\n

Question 17.
\nA building contractor needs three helpers and ten men apply. In how many ways can these selections take place ? [1 Mark, Nov. 2007]
\n(a) 36
\n(b) 15
\n(c) 150
\n(d) 120
\nAnswer:
\n(d) is correct
\nSelection of 3 helpers out of 10
\n= 10<\/sup>C3<\/sub> = \\(\\frac{10 !}{3 ! .(10-3) !}\\)
\n= \\(\\frac{10.9 .8 .7 !}{3.2 .1 .7 !}\\) = 120<\/p>\n

Question 18.
\nThere are three blue balls, four red balls and five green balls. In how many ways can they be arranged in a row ? [1 Mark. Feb. 2008]
\n(a) 26,720
\n(b) 27,720
\n(c) 27,820
\n(d) 26,620
\nAnswer:
\n(b) is Correct
\nNo. of Arrangements = \\(\\frac{12 !}{3 ! .4 ! .5 !}\\)
\n= 27,720<\/p>\n

Question 19.
\nIf C(n, r): C(n, r+1) =1:2 and C (n, r + 1): C (n, r + 2) = 2:3, determine the value of n and r : [1 Mark, Feb. 2008]
\n(a) (14, 4)
\n(b) (12, 4)
\n(c) (14, 6)
\n(d) None
\nAnswer:
\n(a) is correct
\nTricks : Go by choices
\nn = 14 ; r = 4<\/p>\n

\"Basic<\/p>\n

Question 20.
\nSix seats of articled clerks are vacant in a \u2018Chartered Accountant Firm\u2019. How many different batches of candidates can be chosen out of ten candidates ? [1 Mark, June. 2008]
\n(a) 216
\n(b) 210
\n(c) 220
\n(d) None
\nAnswer:
\n(b) is correct<\/p>\n

Question 21.
\nSix persons A, B, C, D, E and F are to be seated at a circular table. In how many ways can this be done, if A must always have either B or C on his right and B must always have either C or D on his right ? [1 Mark, June. 2008]
\n(a) 3
\n(b) 6
\n(c) 12
\n(d) 18
\nAnswer:
\nFrom Qts.
\nB or C is right side of A
\n\u2234 Arrangements are AB, AC,
\nC or D is right side of B.
\n\u2234 Their arrangements are BC or BD
\n\u2234 Arrangements can be made as
\nCase I
\nABC, D,E,F can be arranged as (4 – 1)! = 6
\nCase II
\nABD C, E, F can be arranged as = (4 – 1)! = 6
\nCase III
\nAC, BD, E, F can be arranged as = (4 – 1)! = 6
\nTotal ways to arrange 6 persons = 6 + 6 + 6 = 18<\/p>\n

Question 22.
\nIf n<\/sup>Pr<\/sub> = n<\/sup>Pr+1<\/sub> and n<\/sup>Cr<\/sub> = n<\/sup>Cr-1<\/sub>, then find the value of \u2018n\u2019. [1 Mark, Dec. 2008]
\n(a) 2
\n(b) 3
\n(c) 4
\n(d) 5
\nAnswer:
\n(b), It n<\/sup>Cr<\/sub> = n<\/sup>Cr-1<\/sub>
\nr + r – 1 = n
\nor 2r = n +1
\nr =
\nIt n<\/sup>Pr<\/sub> = n<\/sup>Pr+1<\/sub>
\n\u2234 Go by Heat & Trial Method For option (b)
\nr must be a whole No.
\n\u2234 Put n = 3 \u2234 We get = \\(\\frac{3+1}{2}\\) = 2
\n\u2234 From 1st Condition
\nLHS = n<\/sup>Pr<\/sub> = 3! = 6
\nRHS = n<\/sup>Pr+1<\/sub> = 3<\/sup>P2+1<\/sub>= 3<\/sup>P3<\/sub> = 6
\n\u2234 n = 3 Satisfy it
\n\u2234 (b) is correct<\/p>\n

Question 23.
\nHow many six digit telephone numbers can be formed by using 10 distinct digits? [1 Mark, Dec. 2008]
\n(a) 106<\/sup>
\n(b) 610<\/sup>
\n(c) 10<\/sup>C6<\/sub>
\n(d) 10<\/sup>P6<\/sub>
\nAnswer:
\n(d) is correct
\nTotal No. of digits = 10
\ni.e. they are 0, 1, 2, 3, ……….;9
\nNo. of six digit distinct Telephone
\nNo. being distinct digits being distinct digits
\n= 10<\/sup>P6<\/sub><\/p>\n

Question 24.
\nIn how many ways a committee of 6 members can be formed from a group of 7 boys and 4 girls having at least 2 girls in the committee. [1 Mark, Dec. 2008]
\n(a) 731
\n(b) 137
\n(c) 371
\n(d) 351
\nAnswer:
\n(c) is correct
\n\"Basic
\nTotal = 371<\/p>\n

Question 25.
\nNumber of ways of painting a face of a cube by 6 colours is: [1 Mark, June 2009]
\n(a) 36
\n(b) 6
\n(c) 24
\n(d) 1
\nAnswer:
\n(b) is correct
\nNo. of faces of a cube = 6
\nNo. of colours = 6
\nNo. of ways to point 1 face = 6<\/sup>C1<\/sub> = 6<\/p>\n

Question 26.
\nIf 18<\/sup>Cr<\/sub> = 18<\/sup>Cr+2<\/sub> fmd the value of r<\/sup>C5<\/sub>.
\n(a) 55
\n(b) 50
\n(c) 56
\n(d) None of these
\nAnswer:
\n(c) is correct
\nIf 18<\/sup>Cr<\/sub> = 18<\/sup>Cr+2<\/sub>
\nr + r + 2 = 18
\nor 2r =16 \u2234 r = 8
\n\u2234 r<\/sup>C5<\/sub> = 8<\/sup>C5<\/sub> = 56<\/p>\n

\"Basic<\/p>\n

Question 27.
\n7 books are to be arranged in such a way so that two particular books are always at first and last place. Final the number of arrangements. [1 Mark, June 2009]
\n(a) 60
\n(b) 120
\n(c) 240
\n(d) 480
\nAnswer:
\n(c) is correct
\nArrangement of 2 Particular books = 2! = 2
\ni.e. either at 1st place or at last place
\nArrangement of rest 5 books = 5! = 120
\n\u2234 Total arrangements = 2 \u00d7 120 = 240<\/p>\n

Question 28.
\nFind the number of arrangements in which the letters of the word \u2018MONDAY\u2019 be arranged so that the words thus formed begin with \u2018M\u2019 and do not end with \u2018N\u2019. [1 Mark, June 2009]
\n(a) 720
\n(b) 120
\n(c) 96
\n(d) None
\nAnswer:
\n(c) is correct MONDAY
\nTotal No. of words which begins with M = (6 – 1)! = 120
\nTotal No. of words which begins with M and ends with N
\n= 1 \u00d7 4! \u00d7 1 = 24
\nTotal No. of words which begins with M but do not end with N = 120 – 24 = 96
\nTricks Use box method<\/p>\n

Question 29.
\nIn how many ways can 17 billiard balls be arranged if 7 of them are black, 6 red and 4 white? [1 Mark, June 2009]
\n(a) 4084080
\n(b) 1
\n(c) 8048040
\n(d) None of these
\nAnswer:
\n(a) is correct Total No. of arrangements
\n= \\(\\frac{17 !}{7 ! \\cdot 6 ! .4 !}=\\frac{17 \\cdot 16 \\cdot 15 \\cdot 14 \\cdot 13 \\cdot 12.11 .10 \\cdot 9.8 .7 !}{7 ! \\cdot 6 ! .4 !}\\)
\n= 4084080<\/p>\n

Question 30.
\n(n + 1)! = 20(n – 1)!, Find n. [1 Mark, Dec. 2009]
\n(a) 6
\n(b) 5
\n(c) 4
\n(d) 10
\nAnswer:
\n(c) is correct (n + 1) ! = 20(n – 1)!
\nor(n + 1)n.(n – 1)! = 20 (n – 1)!
\nor (n + 1).n = 5 \u00d7 4
\nn = 4<\/p>\n

Question 31.
\nOut of 4 gents and 6 ladies, a committee is to be formed. Find the number of ways the committee can be formed such that it comprises of at least 2 gents and at least the number of ladies should be double of gents. [1 Mark, Dec. 2009]
\n(a) 94
\n(b) 132
\n(c) 136
\n(d) 104
\nAnswer:
\n(c) is correct
\n\"Basic
\nTotal = 136<\/p>\n

Question 32.
\nIn how many ways can the letters of \u2018REGULATION\u2019 be arranged so that the vowels come at odd places? [1 Mark, Dec. 2009]
\n(a) 14,400
\n(b) 1,144
\n(c) 1,44,252
\n(d) None of these
\nAnswer:
\n(a) is correct
\nREGULATION
\nConsonants = R, G, L, T,N
\nVowels = E, U, A, I, O
\nOdd places are 1, 3, 5, 7, 9
\nArrangement of vowels at these Places = 5<\/sup>P5<\/sub> = 5!= 120
\nConsonants will be arranged at remaining places = 5! = 120
\nTotal No. of arrangements = 120 \u00d7 120 = 14,400<\/p>\n

Question 33.
\nSix points are on a circle. The number of quadrilaterals that can be formed are: [1 Mark, June 2010]
\n(a) 30
\n(b) 360
\n(c) 15
\n(d) None
\nAnswer:
\n(c) is correct
\nNo. of Quadrilaterals
\n6<\/sup>C4<\/sub> = \\(\\frac{6 !}{4 ! .2 !}=\\frac{6.5 .4 !}{4 ! .2 !}\\) = 15<\/p>\n

\"Basic<\/p>\n

Question 34.
\nThe number of ways of arranging 6 boys and 4 girls in a row so that all 4 girls are together is: [1 Mark, June 2010]
\n(a) 6!.4!
\n(b) 2(71.4!)
\n(c) 71.4!
\n(d) 2.(6!.4!)
\nAnswer:
\n(c) is correct
\nNo. of arrangements of 6 boys and 4 girls so that all 4 girls are together = (6 + 1)! .4!
\n= 7!. 4! = 120960.<\/p>\n

Question 35.
\nHow many numbers not exceeding 1000 can be made from the digits 1,2, 3, 4, 5, 6, 7, 8, 9 if repetition is not allowed. [1 Mark, June 2010]
\n(a) 364
\n(b) 585
\n(c) 728
\n(d) 819
\nAnswer:
\n(b) is correct
\nGiven digits = 1, 2, 3, ………… ;9
\nNumbers less than 1000 will be of 1 digit 2 digits and of 3 digits.
\n\u2234 Total No. of Numbers = 9<\/sup>P1<\/sub> + 9<\/sup>P2<\/sub> + 9<\/sup>P3<\/sub>
\n= 9 + 9 \u00d7 8 + 9 \u00d7 8 \u00d7 7 = 585<\/p>\n

Question 36.
\nA garden having 6 tall trees in a row. In how many ways 5 children stand, one in a gap between the trees in order to pose for a photograph? [1 Mark, Dec. 2010]
\n(a) 24
\n(b) 120
\n(c) 720
\n(d) 30
\nAnswer:
\n(b) is correct
\nTreas
\n1 * 2 * 3 * 4 * 5 * 6
\nClearly there will be 5 positions for children = 5<\/sup>P5<\/sub> = 120<\/p>\n

Question 37.
\n15<\/sup>C3<\/sub> + 15<\/sup>C3<\/sub> is equal to: [1 Mark, Dec. 2010]
\n(a) 16<\/sup>C3<\/sub>
\n(b) 30<\/sup>C16<\/sub>
\n(c) 15<\/sup>C16<\/sub>
\n(d) 15<\/sup>C15<\/sub>
\nAnswer:
\n(a) is correct
\n15<\/sup>C3<\/sub> + 15<\/sup>C2<\/sub> = 16<\/sup>C3<\/sub><\/p>\n

Question 38.
\nHow many ways a team of 11 players can be made out of 15 players if one particular player is not to be selected in the team. [1 Mark, Dec. 2010]
\n(a) 364
\n(b) 728
\n(c) 1,001
\n(d) 1,234
\nAnswer:
\nNo .of ways to make a 11 – member teams
\n= 15-1<\/sup>C11<\/sub> = 14<\/sup>C11<\/sub> = \\(\\frac{14 !}{11 ! 3 !}\\)
\n= 364<\/p>\n

Question 39.
\nFind the number of arrangements of 5 things taken out of 12 things, in which one particular thing must always be included. [1 Mark, June 2011]
\n(a) 39,000
\n(b) 37,600
\n(c) 39,600
\n(d) 36,000
\nAnswer:
\n(c) is correct
\nNo. of arrangements of 5 things
\n= 12-1<\/sup>C5-1<\/sub>.5! = 11<\/sup>C4<\/sub>.5!
\n= \\(\\frac{11 !}{4 ! .7 !}\\) \u00d7 120 = 330 \u00d7 120 = 39600<\/p>\n

Question 40.
\nIn how many ways 3 prizes out of 5 can be distributed amongst 3 brothers Equally ? [1 Mark, Dec. 2011]
\n(a) 10
\n(b) 45
\n(c) 60
\n(d) 120
\nAnswer:
\n(c) is correct
\nNo. of ways = 5<\/sup>C1<\/sub>.4<\/sup>C1<\/sub>.3<\/sup>C1<\/sub> = 5.4.3 = 60<\/p>\n

\"Basic<\/p>\n

Question 41.
\nThere are 12 question are to be answered to be Yes or No. How many ways can these be answered ? [1 Mark, Dec. 2011]
\n(a) 1024
\n(b) 2048
\n(c) 4096
\n(d) None
\nAnswer:
\n(c) No. of ways = 212<\/sup> = 4096<\/p>\n

Question 42.
\nThe letters ofthe word VIOLENT are arranged so that the vowels occupy even place only. The number of permutations is: [1 Mark, June. 2012]
\n(a) 144
\n(b) 120
\n(c) 24
\n(d) 72
\nAnswer:
\n(a) Vowels = I, O, E
\nConsonants = V, L, N, T
\nNo. of perms of vowels = 3<\/sup>P3<\/sub> = 3! = 6
\nTotal no. of words = 4! \u00d7 3! = 24 \u00d7 6 = 144<\/p>\n

Question 43.
\nIf n<\/sup>P4<\/sub> = 2o(n<\/sup>P2<\/sub>)then the value of \u2018n\u2019 is ______. [1 Mark, June 2012]
\n(a) -2
\n(b) 7
\n(c) -2 and 7 both
\n(d) None of these
\nAnswer:
\n(b) Note : n is always positive Go by choices.<\/p>\n

Question 44.
\nA man has 3 sons and 6 schools within his reach, in how many ways, he can send them to school, if no two of his sons are to read in the same school ? [1 Mark, Dec. 2012]
\n(a) 6<\/sup>P2<\/sub>
\n(b) 6<\/sup>P3<\/sub>
\n(c) 6<\/sup>P3<\/sub>
\n(d) 36<\/sup>
\nAnswer:
\n(b) No. of ways = 6<\/sup>P3<\/sub><\/p>\n

Question 45.
\nHow many permutations can be formed from the letters of the word \u201cDRAUGHT\u201d, if both vowels may not be separated? [1 Mark, Dec. 2012]
\n(a) 720
\n(b) 1,440
\n(c) 140
\n(d) 1,000
\nAnswer:
\n(b) Total Perms, of DRAUGHT = 6!.2! = 720 \u00d7 2 = 1440.<\/p>\n

Question 46.
\nIf 13<\/sup>C6<\/sub> + 213<\/sup>C5<\/sub> + 13<\/sup>C4<\/sub> = 13<\/sup>Cx<\/sub> then, x = __. [1 Mark, Dec. 2012]
\n(a) 6
\n(b) 7
\n(c) 8
\n(d) 9
\nAnswer:(a)
\n13<\/sup>C6<\/sub> + 213<\/sup>C5<\/sub> + 13<\/sup>C4<\/sub> = 13<\/sup>Cx<\/sub>
\nor 14<\/sup>C6<\/sub> + 14<\/sup>C5<\/sub> = 15<\/sup>Cx<\/sub>
\nor 15<\/sup>C6<\/sub> = 15<\/sup>Cx<\/sub>
\nx = 6<\/p>\n

Question 47.
\nThe total number of shake hands in a group of 10 persons to each other are: [1 Mark, June 2013]
\n(a) 45
\n(b) 54
\n(c) 90
\n(d) 10
\nAnswer:
\n(a) Total no of handshakes = 10<\/sup>C2<\/sub> =45<\/p>\n

Question 48.
\nA regular polygon has 44 diagonals then the No. of sides are: [1 Mark, June 2013]
\n(a) 8
\n(b) 9
\n(c) 10
\n(d) 11
\nAnswer:
\n(d) No. of Diagonals in a polygon of \u201cn\u201d sides = n<\/sup>C2<\/sub> – n = 44
\nTricks : Go by choices.
\nFor.-(a) 8<\/sup>C2<\/sub> – 8 = 28 – 8 \u2260 44
\n(b) 9<\/sup>C2<\/sub> – 9 = 36 – 9 \u2260 44
\n(c) 10<\/sup>C2<\/sub> – 10 = 45 – 10 \u2260 44
\n(d) 11<\/sup>C2<\/sub> – 11 = 55 – 11 = 44
\n\u2234 (d) is correct<\/p>\n

Question 49.
\nIn how many ways the word \u201cARTICLE\u201d can be arranged in a row so that vowels occupy even places? [1 Mark, June 2013]
\n(a) 132
\n(b) 144
\n(c) 72
\n(d) 160
\nAnswer:
\n(b) In word ARTICLE Vowels =A,I,E Positions For vowels = 2, 4, 6
\n\u2234 Vowels can be arranged in 3! = 6 ways
\nRest letters can be arranged in 4! = 24 ways
\n\u2234 Total No. of such arrangements
\n= 6 \u00d7 24 = 144<\/p>\n

Question 50.
\nHow many different words can be formed with the letters of the word \u201cLIBERTY\u201d: [1 Mark, Dec. 2013]
\n(a) 4050
\n(b) 5040
\n(c) 5400
\n(d) 4500
\nAnswer:
\n(b) is correct
\nLIBERTY
\nNo. of words = 7! = 5040<\/p>\n

Question 51.
\nIn how ways can a family consist of 3 children have different birthday in a leap year: [1 Mark, Dec. 2013]
\n(a) 366 \u00d7 365 \u00d7 364
\n(b) 366<\/sup>C6<\/sub>
\n(c) 365<\/sup>C3<\/sub>
\n(d) 366<\/sup>C3<\/sub> – 3
\nAnswer:
\n(a) is correct
\n1 Leap year = 366
\nNo. ofways = 366<\/sup>C1<\/sub>.365<\/sup>C1<\/sub>.364<\/sup>C1<\/sub>
\n= 366 \u00d7 365 \u00d7 364<\/p>\n

Question 52.
\nIf 15<\/sup>C3r<\/sub> = 15<\/sup>Cr+3<\/sub> then r = [1 Mark, Dec. 2013]
\n(a) 2
\n(b) 3
\n(c) 4
\n(d) 5
\nAnswer:
\n(b) is correct
\n15<\/sup>C3r<\/sub> = 15<\/sup>Cr+3<\/sub>
\nEither 3r = r + 3
\n2r = 3 ; So, r = \\(\\frac{3}{2}\\) (In fraction; so invalid)
\nOr 3r + r + 3 = 15
\nor 4r = 12 ; So, r = 3<\/p>\n

\"Basic<\/p>\n

Question 53.
\nIf 6 times the No. of permutations of n items taken 3 at a times is equal to 7 times the No. of permutations of (n-1) items taken 3 at a time then the value of n will be: [1 Mark, June. 2014]
\n(a) 1
\n(b) 9
\n(c) 13
\n(d) 21
\nAnswer:
\n(d) is correct 6,”P3 =7.(n_1)P3 (given)
\n6.n<\/sup>P3<\/sub> = 7.(n-1)<\/sup>C3<\/sub> (given)
\nor 6. \\(\\frac{n !}{(n-3) !}\\) = 7.\\(\\frac{(n-1) !}{(n-1-3) !}\\)
\nor; \\(\\frac{6 \\cdot n \\cdot(n-1) !}{(n-3)(n-4) !}=\\frac{7(n-1) !}{(n-4) !}\\)
\nor \\(\\frac{6 n}{n-3}\\) = 7
\nor 6n = 7n – 21
\nor n = 21
\nTricks : By Calculator (Go by choices)<\/p>\n

Question 54.
\nIf 1000<\/sup>C98<\/sub> = 999<\/sup>C97<\/sub> + x<\/sup>C901<\/sub> then the value of x will be: [1 Mark, June 2014]
\n(a) 999
\n(b) 998
\n(c) 997
\n(d) None
\nAnswer:
\n(a) is correct Tricks : Go by choices
\nFor (a) RHS = 999<\/sup>C97<\/sub> + 999<\/sup>C901<\/sub>
\n= 999<\/sup>C97<\/sub> + 999<\/sup>C98<\/sub> [\u2234 n<\/sup>Cr<\/sub> = n<\/sup>Cn-r<\/sub>]
\n= 1000<\/sup>C98<\/sub> (L.H.S)
\n[\u2235 n<\/sup>Cr<\/sub> + n<\/sup>Cr-1<\/sub> = n+1<\/sup>Cr<\/sub>]<\/p>\n

Question 55.
\n6<\/sup>Pr<\/sub> = 360 then find r. [1 Mark, Dec. 2014]
\n(a) 4
\n(b) 5
\n(c) 6
\n(d) None
\nAnswer:
\n(a) is correct
\nTricks: Go by choices:
\n6<\/sup>Pr<\/sub> = 360
\nLHS at r = 4
\n= 6<\/sup>P4<\/sub> = 6.5.4.3 = 360. R.H.S.<\/p>\n

Question 56.
\nIf 5 books of English 4 books of Tamil and 3 books of Hindi are to be arranged in a single row so that books of same language come together: [1 Mark, Dec. 2014]
\n(a) 1,80,630
\n(b) 1,60,830
\n(c) 1,03,680
\n(d) 1,30,680
\nAnswer:
\n(c) is correct
\nTotal No. of ways so that same language books remain together
\n= 5 !. 4 !. 3 !. 3 !.
\n= 120 \u00d7 24 \u00d7 6 \u00d7 6
\n= 1,03,680<\/p>\n

Question 57.
\n5 Boys and 4 girls are to be seated in row. If the girls occupy even places then the No. of such arrangements: [1 Mark, Dec. 2014]
\n(a) 288
\n(b) 2808
\n(c) 2008
\n(d) 2880
\nAnswer:
\n(d) is correct Total No. of students = 9
\nFor girls positions may be 2,4,6,8,
\nTotal no. of arrangements of girls
\n4<\/sup>P4<\/sub> = 4 1=24
\nFor boys = 5 ! = 120
\nTotal ways = 5!. 4! = 120 \u00d7 24 = 2880<\/p>\n

Question 58.
\nA person has 10 friends of which 6 of them are relatives. He wishes to invite 5 persons so that 3 of them are relatives. In how many ways he can invites? [1 Mark, June 2015]
\n(a) 450
\n(b) 600
\n(c) 120
\n(d) 810
\nAnswer:
\nFriends =10
\nRelatives = 6 ; So, Rest = 4 friends
\nSelection of 5 in which 3 are relatives
\n= 6<\/sup>C3<\/sub>.4<\/sup>C2<\/sub> = 20 \u00d7 6 = 120
\n(c) is correct<\/p>\n

Question 59.
\nA student has 3 books on computer, 3 books on Economics, 5 on Commerce.If these books are to be arranged subject wise then these can be placed on a shelf in the_ _ number, of ways. [1 Mark, June 2015]
\n(a) 25.290
\n(b) 25,920
\n(c) 4,230
\n(d) 4,320
\nAnswer:
\nTotal ways = 3! \u00d7 3! \u00d7 5! \u00d7 3!
\n= 6 \u00d7 6 \u00d7 120 \u00d7 6
\n= 25920<\/p>\n

Question 60.
\nThe number of 4 digit numbers that can be formed from seven digits 1, 2, 3, 5, 7, 8, 9 such that no digit being repeated in any number,Which are greater than 3000 are: [1 Mark, June 2015]
\n(a) 120
\n(b) 480
\n(c) 600
\n(d) 840
\nAnswer:
\nFor Detail see Quicker BMLRS<\/p>\n\n\n\n
5<\/td>\n6<\/td>\n5<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

At 1st place 3, 5, 7, 8, 9 these 5 digits are suitable.
\nSo, permutation of 1 st place = 5
\nOne of them will be used at that place.
\nRemaining 6 digits are suitable for next place.
\nSimilarly doing as above,
\nTotal ways = 5.6.5.4 = 600
\n(c) is correct.<\/p>\n

Question 61.
\nA question paper consist 10 questions, 6 in math and 4 in stats. Find outnumber of ways to solve question paper if at least one question is to be attempted from each section. [1 Mark, Dec. 2015]
\n(a) 1024
\n(b) 950
\n(c) 945
\n(d) 1022
\nAnswer:
\n(c) is correct
\nNo. of ways to attempt at least one from each = (26<\/sup> – 1)(24<\/sup> – 1) = 945.<\/p>\n

Question 62.
\nThere are 6 gents and 4 ladies. A committee of 5 is to be formed if it include at least two ladies. [1 Mark, Dec. 2015]
\n(a) 64
\n(b) 162
\n(c) 102
\n(d) 186
\nAnswer:
\n(d) is correct
\n\"Basic<\/p>\n

Question 63.
\nn<\/sup>Pr<\/sub> = 720 and n<\/sup>Cr<\/sub> =120 ,Find r? [1 Mark, Dec. 2015]
\n(a) 6
\n(b) 4
\n(c) 3
\n(d) 2
\nAnswer:
\n(c) is correct n<\/sup>Pr<\/sub> = n<\/sup>Cr<\/sub> r! = 120
\n120.r! = 720
\nr! = 6 = 3!
\n\u2234 r = 3<\/p>\n

\"Basic<\/p>\n

Question 64.
\nThere are 10 students in a class, including 3 girls. The number of ways to arrange them in a row, when any two girls out of them never come together: [1 Mark, June 2016]
\n(a) 8<\/sup>P3<\/sub> \u00d7 7!
\n(b) 3<\/sup>P3<\/sub> \u00d7 7!
\n(c) 8<\/sup>P3<\/sub> \u00d7 10!
\n(d) None
\nAnswer:
\n(a)
\nRest Students = 7
\nTotal no. of permutations of rest 7 stds. = 7 !
\n\"Basic
\nTotal no. of places for girls = 7+1 = 8
\nTotal ways to arrange girls = 8<\/sup>P3<\/sub>
\nTotal No. of required ways = 8<\/sup>P3<\/sub> . (7!)<\/p>\n

Question 65.
\nIn how many ways can a selection of 6 out of 4 teachers and 8 students be done so as to include atleast two teachers ?
\n(a) 220
\n(b) 672
\n(c) 896
\n(d) 968
\nAnswer:
\n(b)
\n\"Basic<\/p>\n

Question 66.
\nThe maximum number of points of inter section of 10 circles will be: [1 Mark, June 2016]
\n(a) 2
\n(b) 20
\n(c) 90
\n(d) 180
\nAnswer:
\n(c)
\nTo intersect 2 circles out of 10 are required
\n= 10<\/sup>C2<\/sub> \u00d7 2 = 45 \u00d7 2 = 90 Nos. of ways
\n(Each circle will cut at 2 points)<\/p>\n

Question 67.
\nHow many numbers between 1,000 and 10,000 can be formed with the digits 1,2, 3, 4, 5, 6. [1 Mark, Dec. 2016]
\n(a) 720
\n(b) 360
\n(c) 120
\n(d) 60
\nAnswer:
\n(b) is correct.
\nTotal no. of numbers = 6<\/sup>P4<\/sub> = 6.5.4.3 = 360<\/p>\n

Question 68.
\nIf n+1<\/sup>Cr+1<\/sub>: n<\/sup>Cr<\/sub>: n-1<\/sup>C2<\/sub> = 8 : 3 : 1; then find the value of n. [1 Mark, Dec. 2016]
\n(a) 14
\n(b) 15
\n(c) 16
\n(d) 17
\nAnswer:
\n(b) is correct.
\nFormula , n<\/sup>Cr<\/sub> : n-1<\/sup>Cr-1<\/sub> = \\(\\frac{n}{r}=\\frac{3}{1}\\); So, n = 3r
\n& n+1<\/sup>Cr+1<\/sub> : n<\/sup>Cr<\/sub> = \\(\\frac{n+1}{r+1}=\\frac{8}{3}\\)
\nSo, 3n + 3 = 8r + 8
\nor, 3 \u00d7 3r + 3 = 8r + 8
\nor, r = 8 – 3 =5
\nHence, n = 3r = 3 \u00d7 5 = 15<\/p>\n

Question 69.
\nIn how many ways 4 members can occupy 9 vacant seats in a row: [1 Mark, Dec. 2016]
\n(a) 3204
\n(b) 3024
\n(c) 49<\/sup>
\n(d) 94<\/sup>
\nAnswer:
\n(b) is correct
\nTotal ways = 9<\/sup>P4<\/sub> = 9.8.7.6 = 3024<\/p>\n

\"Basic<\/p>\n

Question 70.
\nThe number of arrangements that can be formed from the letters of the word \u201cALLAHABAD\u201d. [1 Mark, June 2017]
\n(a) 7560
\n(b) 3780
\n(c) 30240
\n(d) 15320
\nAnswer:
\n\\(\\frac{9 !}{(4 !) \\cdot(2 !)}=\\frac{9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5(4 !)}{(4 !) \\cdot 2 \\cdot 1 .}\\) = 7560
\nOption (a) is correct<\/p>\n

Question 71.
\nIf 10<\/sup>C3<\/sub> + 2.10<\/sup>C4<\/sub> + 10<\/sup>C5<\/sub> = n<\/sup>C5<\/sub> then the value of n = .[1 Mark, June 2017]
\n(a) 10
\n(b) 11
\n(c) 12
\n(d) 13
\nAnswer:
\n10<\/sup>C3<\/sub> + 2.10<\/sup>C4<\/sub> + 10<\/sup>C5<\/sub> = n<\/sup>C5<\/sub>
\n= 10<\/sup>C3<\/sub> + 10<\/sup>C4<\/sub> + 10<\/sup>C4<\/sub> + 10<\/sup>C5<\/sub> = n<\/sup>C5<\/sub>
\nor; 11<\/sup>C4<\/sub> + 11<\/sup>C5<\/sub> = n<\/sup>C5<\/sub>
\nor; 12<\/sup>C5<\/sub> = n<\/sup>C5<\/sub> \u21d2 n = 12
\nOption (c) is correct<\/p>\n

Question 72.
\nThe number of parallelograms that can be formed by a set of 6 parallel lines intersected by the another set of 4 parallel lines is: [1 Mark, June 2017]
\n(a) 360
\n(b) 90
\n(c) 180
\n(d) 45
\nAnswer:
\nNo. of parallelograms = 6<\/sup>C2<\/sub> . 4<\/sup>C2<\/sub>
\n= 15 \u00d7 6 = 90
\nOption (b) is correct<\/p>\n

Question 73.
\nIf n<\/sup>P13<\/sub> : (n+1)<\/sup>P12<\/sub> = 3 : 4 then \u2018n\u2019 is : [1 Mark, Dec. 2017]
\n(a) 13
\n(b) 15
\n(c) 18
\n(d) 31
\nAnswer:
\n(b) is correct
\nTricks : Go by choices (Test from beginning)
\n\"Basic<\/p>\n

\"Basic<\/p>\n

Question 74.
\nIn how many ways that 3 commerce books, 3 computer books and 5 economics books be arranged along a row, so that books of same subjects are come together is: [1 Mark, Dec. 2017]
\n(a) 29,950
\n(b) 25,940
\n(c) 25,920
\n(d) None of these
\nAnswer:
\n(c)
\nFrom Qts. ; Total no. of ways
\n= (3!) . (3!) . (5!) . (3!)
\n= 6 \u00d7 6 \u00d7 120 \u00d7 6
\n= 25,920.<\/p>\n

Question 75.
\nIf 12<\/sup>C3<\/sub> + 2.12<\/sup>C4<\/sub> + 12<\/sup>C5<\/sub> = 12<\/sup>C5<\/sub>,The value of x : [1 Mark, June 2018]
\n(a) 3 or 5
\n(b) 5 or 9
\n(c) 7 or 1
\n(d) 9 or 12
\nAnswer:
\n(b)
\n12<\/sup>C3<\/sub> + 2.12<\/sup>C4<\/sub> + 12<\/sup>C5<\/sub> = 12<\/sup>C5<\/sub>
\n\u21d2 12<\/sup>C3<\/sub> + 12<\/sup>C4<\/sub> + 12<\/sup>C4<\/sub> + 12<\/sup>C5<\/sub> = 14<\/sup>Cx<\/sub>
\n\u21d2 13<\/sup>C4<\/sub> + 13<\/sup>C5<\/sub> = 14<\/sup>Cx<\/sub>
\n\u21d2 14<\/sup>C5<\/sub> = 14<\/sup>C14-5<\/sub> = 14<\/sup>Cx<\/sub>
\n\u21d2 x = 14 – 5 = 9
\nx = 5 or 9<\/p>\n

Question 76.
\nThe number of ways in which a man can invite one or more of his 7 friends to dinner is
\n(a) 64
\n(b) 128
\n(c) 127
\n(d) 63
\nAnswer:
\n(c)
\nNo. of ways to invite at least one friend = 27 – 1
\n= 128 – 1
\n= 127.<\/p>\n

Question 77.
\nThe number of words from the letters of the word BHARAT, in which B and H will never come together, is:[1 Mark, Nov. 2018]
\n(a) 120
\n(b) 360
\n(c) 240
\n(d) None
\nAnswer:
\n(c)
\nUse Gap rule.
\n*A * R * A * T *
\nTotal words = \\(\\frac{4 !}{2 !}\\) \u00d7 5<\/sup>P2<\/sub>
\n= \\(\\frac{24}{2}\\) \u00d7 5 \u00d7 4 = 240<\/p>\n

Question 78.
\nThe value of N in \\(\\frac{1}{7 !}+\\frac{1}{8 !}=\\frac{N}{9 !}\\) is [1 Mark, Nov 2018]
\n(a) 81
\n(b) 64
\n(c) 78
\n(d) 89
\nAnswer:
\n(a)
\n\"Basic
\n\u21d2 N = 81<\/p>\n

Question 79.
\nIf n<\/sup>Pr<\/sub> = 720 and n<\/sup>Cr<\/sub> = 120 then r is: [1 Mark, Nov. 2018]
\n(a) 4
\n(b) 5
\n(c) 3
\n(d) 6
\nAnswer:
\n(c)
\nn<\/sup>Pr<\/sub> = n<\/sup>Cr<\/sub>.r!
\nor; 720 = 120 (r!)
\nor 6 = r! \u21d2 r! = 3!
\nr = 3<\/p>\n

\"Basic<\/p>\n

Question 80.
\nA bag contains 4 red, 3 black and 2 white balls. In how many ways 3 balls can be drawn from this bag so that they include at least one black ball ? [1 Mark, Nov. 2018]
\n(a) 46
\n(b) 64
\n(c) 86
\n(d) None
\nAnswer:
\n(b)
\n\"Basic<\/p>\n","protected":false},"excerpt":{"rendered":"

This Basic Concepts of Permutations and Combinations \u2013 CA Foundation Maths Study Material\u00a0is designed strictly as per the latest syllabus and exam pattern. Basic Concepts of Permutations and Combinations \u2013 CA Foundation Maths Study Material Permutations and combinations : There are various ways in which objects from a set may be selected, generally without replacement, …<\/p>\n

Basic Concepts of Permutations and Combinations \u2013 CA Foundation Maths Study Material<\/span> Read More »<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[160],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/posts\/13265"}],"collection":[{"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/comments?post=13265"}],"version-history":[{"count":3,"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/posts\/13265\/revisions"}],"predecessor-version":[{"id":13288,"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/posts\/13265\/revisions\/13288"}],"wp:attachment":[{"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/media?parent=13265"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/categories?post=13265"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gstguntur.com\/wp-json\/wp\/v2\/tags?post=13265"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}