Basic Concepts of Permutations and Combinations – CA Foundation Maths Study Material

This Basic Concepts of Permutations and Combinations – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

Basic Concepts of Permutations and Combinations – CA Foundation Maths Study Material

Permutations and combinations :
There 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¬tation when the order of selection is a factor and this selection of subsets is called a Combination when order is not a factor.
The 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

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 “3 permute 2.” 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
nPr = $$\frac{n !}{(n-r) !}$$
3P2 = $$\frac{3 !}{(3-2) !}$$ = 3! = 3.2.1 = 6
For Combinations, r objects are selected from a set of n objects to produce subsets without ordering. Contrasting the previous permutation example with the corre¬sponding combination, the AB and BA subsets are no longer distinct selections; by eliminating such cases there remain only 3 different possible subsets—AB, AC, BC.

The number of such subsets is denoted by nCr, read “n choose r” & is formulated as nCr = $$\frac{n !}{r !(n-r) !}$$

Fundamental Rule of Counting
Ex- 7 buses
CP Laxmi Nagar
5 Autoes
There 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.
∴ Total ways to reach Laxmi Nagar from CP
= 7 + 5 = 12 ways.
Rule: If there are two alternative jobs,one can be done by “p” ways and another by “q” ways, either of these two jobs can be done by (p + q) ways.

(ii) Multiplication (Product) Rule:
If one things is done by “p” different ways and when it has been done a second thing can be done by “q” different ways then total number of ways of doing both things simultaneously
= p × q. ways

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.
∴ Total No. of ways to reach Jaipur and come back
= 7 × 6 = 42 ways

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.
Answer:Total No. of ways = 5 × 4 × 6 = 120
Note:
(i) In case of either or; or; one of them; atleast one; or options then use addition rule
(ii) In case of AND ; then use product rule.

Previous Year Exam Questions

Question 1.
The 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]
(a) 185
(b) 175
(c) 115
(d) 105
(a) is Correct
No. of triangles = 12C37C3
= $$\frac{12 !}{3 ! 9 !}-\frac{7 !}{3 ! 4 !}$$ = 220 – 35
= 185

Question 2.
A 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]
(a) 6,15,800
(b) 46,800
(c) 7,19,500
(d) 4,10,800
(b) No. of code Words = 26P2 × 9P2
= $$\frac{26 \cdot 25 \cdot 24 !}{24 !} \times \frac{9 !}{7 !}$$
= 650 × 72 = 46,800

Question 3.
A 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]
(a) 41
(b) 51
(c) 61
(d) 71
(a) is correct

Total Combinations = 6+15 + 20 = 41

Question 4.
An 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]
(a) 35
(b) 175
(c) 210
(d) 420
(d) is correct 12 Qts.

Total No. of ways to select Qts.
= 7C3.5C5 + 7C4.5C4 + 7C5.5C3
= 35 × 1 + 35 × 5 + 21 × 10
= 35 + 175 + 210 = 420

Question 5.
A Supreme Court Bench consists of 5 judges. In how many ways, the bench can give a majority division ? [1 Mark, Feb. 2007]
(a) 10
(b) 5
(c) 15
(d) 16
(d) is correct

No. of ways to give majority decision
= 5C3 + 5C4 + 5C3
= 10 + 5 + 1
= 16

Question 6.
Given: P (7, k) = 60 P(7, k – 3). Then: [1 Mark, Feb. 2007]
(a) k = 9
(b) k = 8
(c) k = 5
(d) k = 0
(c) is Correct
P(7, K) = 60.p(7, k-3)

or (10 – k) (9 – k) (8 – k) = 60
or (10 – k) (0 – k) (8 – k) = 5.4.3
[make 3 factors of 60 so that their differences are 1.]
Comparing on both sides ; we get
10 – k = 5
or k= 10 – 5 = 5
∴ k = 5

Question 7.
The 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]
(a) (n – 2) × (n – 1)!
(b) (n – 2) × (n + l)!
(c) (n – 1) × (n + 1)!
(d) (n – 2) × (n + 2)!
(a) is correct
Total No. of ways to arrange book as Qts.
= n! – (n – 2 + 1)! .2 !
= n! – (n – 1)!.2!
= n (n – 1)! – 2 (n – 1)!
= (n – 1)!. (n – 2)
= (n – 2). (n – 1)!

Question 8.
In 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]
(a) 576
(b) 476
(c) 376
(d) 276
(a) is correct Consonants = F, L, R
Vowels = A, I, U, E
Total letters = 7
There will be 7 Positions of letters.
Odd positions are = 1, 3, 5, 7
Even positions are = 2, 4, 6
From Questions
No. of Words = 4P3 .4! = 4! × 4!
= 24 × 24 = 576

Question 9.
Five 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]
(a) 10
(b) 1
(c) 3
(d) None of these
(b) is correct No. of bulbs = 5
Defective bulbs = 3
Non-Defective bulbs = 2
No. of Point in the room = 2
No. of ways to light the darkroom
= 3C1.2C1 + 3C0.2C2
= 3 × 2 + 1 × 1
= 6 + 1 = 7

Question 10.
In 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]
(a) 164
(b) 174
(c) 144
(d) 154
(c) is correct.
No. of ways to sit 4 men around a table = (4 – 1)! = 6
No. of ways to sit 4 women = 4! = 24
∴ Total Ways to arrange = 6 × 24 = 144

Question 11.
The value of $$\sum_{r=1}^5$$ 5Cr is : [1 Mark, May 2007]
(a) 29
(b) 31
(c) 35
(d) 26
(b) is correct
$$\sum_{r=1}^5$$ 5Cr = 25 -1 = 31

Question 12.
If 6Pr = 24 6Cr, then find r : [1 Mark, Aug. 2007]
(a) 4
(b) 6
(c) 2
(d) 1
(a) is correct
6Pr = 24.6Cr
or $$\frac{6 !}{(6-r) !}=$$ = 24.$$\frac{6 !}{r !(6-r) !}$$
or r! = 24 or r! = 4! r = 4

Question 13.
Find the number of combinations of the letters of the word COLLEGE taken four together: [1 Mark, Aug. 2007]
(a) 18
(b) 16
(c) 20
(d) 26
(a) is correct
Letters are C, O, L, L, E, E, G
Case I 2 same & 2 same letters = LLEE = Only 1 combination
Case II LL and 2 out of C, O, E, G = 1 × 4C2 = 6
EE and any 2 out of C, O, L, G = 1 × 4C2 = 6
Case III All different letters = Any 4 out of C, O, L, E, G = 5C4 = 5
Total No. of combinations = 1 + (6 + 6) + 5 = 18

Question 14.
How many words can be formed with the letters of the word ‘ORIENTAL’ so that A and E always occupy odd places: [1 Mark, Aug. 2007]
(a) 540
(b) 8640
(c) 8460
(d) 8450
(b) is correct
As Qts. No. of Words = 4P2 × 6!
(Because there are 4 Positions 1, 3, 5, 7 for A & E)
= $$\frac{4 !}{(4-2) !}$$ × 720 = 12 × 720 = 8640

Question 15.
If 1000C98 = 999C97 + xC901, find x : [1 Mark, Nov. 2007]
(a) 999
(b) 998
(c) 997
(d) 1000
(a) is correct
Tricks : Go by choices
For (a) RHS = 999C97 + 999C901 (If x = 999 )
= 999C97 + 999C98 = 1000C98 = LHS

Question 16.
How many numbers greater than a million can be formed with the digits 4, 5, 5, 0, 4, 5, 3? [Nov. 2007]
(a) 260
(b) 360
(c) 280
(d) 380
(b) is correct
No. of Numbers = $$\frac{7 !}{3 ! .2 !}$$ – 1 × $$\frac{6 !}{3 ! .2 !}$$
[Zero is fixed at 1st positions with permutation 1]
= 420 – 60 = 360

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

Question 18.
There 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]
(a) 26,720
(b) 27,720
(c) 27,820
(d) 26,620
(b) is Correct
No. of Arrangements = $$\frac{12 !}{3 ! .4 ! .5 !}$$
= 27,720

Question 19.
If 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]
(a) (14, 4)
(b) (12, 4)
(c) (14, 6)
(d) None
(a) is correct
Tricks : Go by choices
n = 14 ; r = 4

Question 20.
Six seats of articled clerks are vacant in a ‘Chartered Accountant Firm’. How many different batches of candidates can be chosen out of ten candidates ? [1 Mark, June. 2008]
(a) 216
(b) 210
(c) 220
(d) None
(b) is correct

Question 21.
Six 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]
(a) 3
(b) 6
(c) 12
(d) 18
From Qts.
B or C is right side of A
∴ Arrangements are AB, AC,
C or D is right side of B.
∴ Their arrangements are BC or BD
∴ Arrangements can be made as
Case I
ABC, D,E,F can be arranged as (4 – 1)! = 6
Case II
ABD C, E, F can be arranged as = (4 – 1)! = 6
Case III
AC, BD, E, F can be arranged as = (4 – 1)! = 6
Total ways to arrange 6 persons = 6 + 6 + 6 = 18

Question 22.
If nPr = nPr+1 and nCr = nCr-1, then find the value of ‘n’. [1 Mark, Dec. 2008]
(a) 2
(b) 3
(c) 4
(d) 5
(b), It nCr = nCr-1
r + r – 1 = n
or 2r = n +1
r =
It nPr = nPr+1
∴ Go by Heat & Trial Method For option (b)
r must be a whole No.
∴ Put n = 3 ∴ We get = $$\frac{3+1}{2}$$ = 2
∴ From 1st Condition
LHS = nPr = 3! = 6
RHS = nPr+1 = 3P2+1= 3P3 = 6
∴ n = 3 Satisfy it
∴ (b) is correct

Question 23.
How many six digit telephone numbers can be formed by using 10 distinct digits? [1 Mark, Dec. 2008]
(a) 106
(b) 610
(c) 10C6
(d) 10P6
(d) is correct
Total No. of digits = 10
i.e. they are 0, 1, 2, 3, ……….;9
No. of six digit distinct Telephone
No. being distinct digits being distinct digits
= 10P6

Question 24.
In 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]
(a) 731
(b) 137
(c) 371
(d) 351
(c) is correct

Total = 371

Question 25.
Number of ways of painting a face of a cube by 6 colours is: [1 Mark, June 2009]
(a) 36
(b) 6
(c) 24
(d) 1
(b) is correct
No. of faces of a cube = 6
No. of colours = 6
No. of ways to point 1 face = 6C1 = 6

Question 26.
If 18Cr = 18Cr+2 fmd the value of rC5.
(a) 55
(b) 50
(c) 56
(d) None of these
(c) is correct
If 18Cr = 18Cr+2
r + r + 2 = 18
or 2r =16 ∴ r = 8
rC5 = 8C5 = 56

Question 27.
7 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]
(a) 60
(b) 120
(c) 240
(d) 480
(c) is correct
Arrangement of 2 Particular books = 2! = 2
i.e. either at 1st place or at last place
Arrangement of rest 5 books = 5! = 120
∴ Total arrangements = 2 × 120 = 240

Question 28.
Find the number of arrangements in which the letters of the word ‘MONDAY’ be arranged so that the words thus formed begin with ‘M’ and do not end with ‘N’. [1 Mark, June 2009]
(a) 720
(b) 120
(c) 96
(d) None
(c) is correct MONDAY
Total No. of words which begins with M = (6 – 1)! = 120
Total No. of words which begins with M and ends with N
= 1 × 4! × 1 = 24
Total No. of words which begins with M but do not end with N = 120 – 24 = 96
Tricks Use box method

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

Question 30.
(n + 1)! = 20(n – 1)!, Find n. [1 Mark, Dec. 2009]
(a) 6
(b) 5
(c) 4
(d) 10
(c) is correct (n + 1) ! = 20(n – 1)!
or(n + 1)n.(n – 1)! = 20 (n – 1)!
or (n + 1).n = 5 × 4
n = 4

Question 31.
Out 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]
(a) 94
(b) 132
(c) 136
(d) 104
(c) is correct

Total = 136

Question 32.
In how many ways can the letters of ‘REGULATION’ be arranged so that the vowels come at odd places? [1 Mark, Dec. 2009]
(a) 14,400
(b) 1,144
(c) 1,44,252
(d) None of these
(a) is correct
REGULATION
Consonants = R, G, L, T,N
Vowels = E, U, A, I, O
Odd places are 1, 3, 5, 7, 9
Arrangement of vowels at these Places = 5P5 = 5!= 120
Consonants will be arranged at remaining places = 5! = 120
Total No. of arrangements = 120 × 120 = 14,400

Question 33.
Six points are on a circle. The number of quadrilaterals that can be formed are: [1 Mark, June 2010]
(a) 30
(b) 360
(c) 15
(d) None
(c) is correct
6C4 = $$\frac{6 !}{4 ! .2 !}=\frac{6.5 .4 !}{4 ! .2 !}$$ = 15

Question 34.
The 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]
(a) 6!.4!
(b) 2(71.4!)
(c) 71.4!
(d) 2.(6!.4!)
(c) is correct
No. of arrangements of 6 boys and 4 girls so that all 4 girls are together = (6 + 1)! .4!
= 7!. 4! = 120960.

Question 35.
How 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]
(a) 364
(b) 585
(c) 728
(d) 819
(b) is correct
Given digits = 1, 2, 3, ………… ;9
Numbers less than 1000 will be of 1 digit 2 digits and of 3 digits.
∴ Total No. of Numbers = 9P1 + 9P2 + 9P3
= 9 + 9 × 8 + 9 × 8 × 7 = 585

Question 36.
A 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]
(a) 24
(b) 120
(c) 720
(d) 30
(b) is correct
Treas
1 * 2 * 3 * 4 * 5 * 6
Clearly there will be 5 positions for children = 5P5 = 120

Question 37.
15C3 + 15C3 is equal to: [1 Mark, Dec. 2010]
(a) 16C3
(b) 30C16
(c) 15C16
(d) 15C15
(a) is correct
15C3 + 15C2 = 16C3

Question 38.
How 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]
(a) 364
(b) 728
(c) 1,001
(d) 1,234
No .of ways to make a 11 – member teams
= 15-1C11 = 14C11 = $$\frac{14 !}{11 ! 3 !}$$
= 364

Question 39.
Find 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]
(a) 39,000
(b) 37,600
(c) 39,600
(d) 36,000
(c) is correct
No. of arrangements of 5 things
= 12-1C5-1.5! = 11C4.5!
= $$\frac{11 !}{4 ! .7 !}$$ × 120 = 330 × 120 = 39600

Question 40.
In how many ways 3 prizes out of 5 can be distributed amongst 3 brothers Equally ? [1 Mark, Dec. 2011]
(a) 10
(b) 45
(c) 60
(d) 120
(c) is correct
No. of ways = 5C1.4C1.3C1 = 5.4.3 = 60

Question 41.
There are 12 question are to be answered to be Yes or No. How many ways can these be answered ? [1 Mark, Dec. 2011]
(a) 1024
(b) 2048
(c) 4096
(d) None
(c) No. of ways = 212 = 4096

Question 42.
The letters ofthe word VIOLENT are arranged so that the vowels occupy even place only. The number of permutations is: [1 Mark, June. 2012]
(a) 144
(b) 120
(c) 24
(d) 72
(a) Vowels = I, O, E
Consonants = V, L, N, T
No. of perms of vowels = 3P3 = 3! = 6
Total no. of words = 4! × 3! = 24 × 6 = 144

Question 43.
If nP4 = 2o(nP2)then the value of ‘n’ is ______. [1 Mark, June 2012]
(a) -2
(b) 7
(c) -2 and 7 both
(d) None of these
(b) Note : n is always positive Go by choices.

Question 44.
A 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]
(a) 6P2
(b) 6P3
(c) 6P3
(d) 36
(b) No. of ways = 6P3

Question 45.
How many permutations can be formed from the letters of the word “DRAUGHT”, if both vowels may not be separated? [1 Mark, Dec. 2012]
(a) 720
(b) 1,440
(c) 140
(d) 1,000
(b) Total Perms, of DRAUGHT = 6!.2! = 720 × 2 = 1440.

Question 46.
If 13C6 + 213C5 + 13C4 = 13Cx then, x = __. [1 Mark, Dec. 2012]
(a) 6
(b) 7
(c) 8
(d) 9
13C6 + 213C5 + 13C4 = 13Cx
or 14C6 + 14C5 = 15Cx
or 15C6 = 15Cx
x = 6

Question 47.
The total number of shake hands in a group of 10 persons to each other are: [1 Mark, June 2013]
(a) 45
(b) 54
(c) 90
(d) 10
(a) Total no of handshakes = 10C2 =45

Question 48.
A regular polygon has 44 diagonals then the No. of sides are: [1 Mark, June 2013]
(a) 8
(b) 9
(c) 10
(d) 11
(d) No. of Diagonals in a polygon of “n” sides = nC2 – n = 44
Tricks : Go by choices.
For.-(a) 8C2 – 8 = 28 – 8 ≠ 44
(b) 9C2 – 9 = 36 – 9 ≠ 44
(c) 10C2 – 10 = 45 – 10 ≠ 44
(d) 11C2 – 11 = 55 – 11 = 44
∴ (d) is correct

Question 49.
In how many ways the word “ARTICLE” can be arranged in a row so that vowels occupy even places? [1 Mark, June 2013]
(a) 132
(b) 144
(c) 72
(d) 160
(b) In word ARTICLE Vowels =A,I,E Positions For vowels = 2, 4, 6
∴ Vowels can be arranged in 3! = 6 ways
Rest letters can be arranged in 4! = 24 ways
∴ Total No. of such arrangements
= 6 × 24 = 144

Question 50.
How many different words can be formed with the letters of the word “LIBERTY”: [1 Mark, Dec. 2013]
(a) 4050
(b) 5040
(c) 5400
(d) 4500
(b) is correct
LIBERTY
No. of words = 7! = 5040

Question 51.
In how ways can a family consist of 3 children have different birthday in a leap year: [1 Mark, Dec. 2013]
(a) 366 × 365 × 364
(b) 366C6
(c) 365C3
(d) 366C3 – 3
(a) is correct
1 Leap year = 366
No. ofways = 366C1.365C1.364C1
= 366 × 365 × 364

Question 52.
If 15C3r = 15Cr+3 then r = [1 Mark, Dec. 2013]
(a) 2
(b) 3
(c) 4
(d) 5
(b) is correct
15C3r = 15Cr+3
Either 3r = r + 3
2r = 3 ; So, r = $$\frac{3}{2}$$ (In fraction; so invalid)
Or 3r + r + 3 = 15
or 4r = 12 ; So, r = 3

Question 53.
If 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]
(a) 1
(b) 9
(c) 13
(d) 21
(d) is correct 6,”P3 =7.(n_1)P3 (given)
6.nP3 = 7.(n-1)C3 (given)
or 6. $$\frac{n !}{(n-3) !}$$ = 7.$$\frac{(n-1) !}{(n-1-3) !}$$
or; $$\frac{6 \cdot n \cdot(n-1) !}{(n-3)(n-4) !}=\frac{7(n-1) !}{(n-4) !}$$
or $$\frac{6 n}{n-3}$$ = 7
or 6n = 7n – 21
or n = 21
Tricks : By Calculator (Go by choices)

Question 54.
If 1000C98 = 999C97 + xC901 then the value of x will be: [1 Mark, June 2014]
(a) 999
(b) 998
(c) 997
(d) None
(a) is correct Tricks : Go by choices
For (a) RHS = 999C97 + 999C901
= 999C97 + 999C98 [∴ nCr = nCn-r]
= 1000C98 (L.H.S)
[∵ nCr + nCr-1 = n+1Cr]

Question 55.
6Pr = 360 then find r. [1 Mark, Dec. 2014]
(a) 4
(b) 5
(c) 6
(d) None
(a) is correct
Tricks: Go by choices:
6Pr = 360
LHS at r = 4
= 6P4 = 6.5.4.3 = 360. R.H.S.

Question 56.
If 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]
(a) 1,80,630
(b) 1,60,830
(c) 1,03,680
(d) 1,30,680
(c) is correct
Total No. of ways so that same language books remain together
= 5 !. 4 !. 3 !. 3 !.
= 120 × 24 × 6 × 6
= 1,03,680

Question 57.
5 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]
(a) 288
(b) 2808
(c) 2008
(d) 2880
(d) is correct Total No. of students = 9
For girls positions may be 2,4,6,8,
Total no. of arrangements of girls
4P4 = 4 1=24
For boys = 5 ! = 120
Total ways = 5!. 4! = 120 × 24 = 2880

Question 58.
A 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]
(a) 450
(b) 600
(c) 120
(d) 810
Friends =10
Relatives = 6 ; So, Rest = 4 friends
Selection of 5 in which 3 are relatives
= 6C3.4C2 = 20 × 6 = 120
(c) is correct

Question 59.
A 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]
(a) 25.290
(b) 25,920
(c) 4,230
(d) 4,320
Total ways = 3! × 3! × 5! × 3!
= 6 × 6 × 120 × 6
= 25920

Question 60.
The 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]
(a) 120
(b) 480
(c) 600
(d) 840
For Detail see Quicker BMLRS

 5 6 5 4

At 1st place 3, 5, 7, 8, 9 these 5 digits are suitable.
So, permutation of 1 st place = 5
One of them will be used at that place.
Remaining 6 digits are suitable for next place.
Similarly doing as above,
Total ways = 5.6.5.4 = 600
(c) is correct.

Question 61.
A 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]
(a) 1024
(b) 950
(c) 945
(d) 1022
(c) is correct
No. of ways to attempt at least one from each = (26 – 1)(24 – 1) = 945.

Question 62.
There 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]
(a) 64
(b) 162
(c) 102
(d) 186
(d) is correct

Question 63.
nPr = 720 and nCr =120 ,Find r? [1 Mark, Dec. 2015]
(a) 6
(b) 4
(c) 3
(d) 2
(c) is correct nPr = nCr r! = 120
120.r! = 720
r! = 6 = 3!
∴ r = 3

Question 64.
There 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]
(a) 8P3 × 7!
(b) 3P3 × 7!
(c) 8P3 × 10!
(d) None
(a)
Rest Students = 7
Total no. of permutations of rest 7 stds. = 7 !

Total no. of places for girls = 7+1 = 8
Total ways to arrange girls = 8P3
Total No. of required ways = 8P3 . (7!)

Question 65.
In how many ways can a selection of 6 out of 4 teachers and 8 students be done so as to include atleast two teachers ?
(a) 220
(b) 672
(c) 896
(d) 968
(b)

Question 66.
The maximum number of points of inter section of 10 circles will be: [1 Mark, June 2016]
(a) 2
(b) 20
(c) 90
(d) 180
(c)
To intersect 2 circles out of 10 are required
= 10C2 × 2 = 45 × 2 = 90 Nos. of ways
(Each circle will cut at 2 points)

Question 67.
How many numbers between 1,000 and 10,000 can be formed with the digits 1,2, 3, 4, 5, 6. [1 Mark, Dec. 2016]
(a) 720
(b) 360
(c) 120
(d) 60
(b) is correct.
Total no. of numbers = 6P4 = 6.5.4.3 = 360

Question 68.
If n+1Cr+1: nCr: n-1C2 = 8 : 3 : 1; then find the value of n. [1 Mark, Dec. 2016]
(a) 14
(b) 15
(c) 16
(d) 17
(b) is correct.
Formula , nCr : n-1Cr-1 = $$\frac{n}{r}=\frac{3}{1}$$; So, n = 3r
& n+1Cr+1 : nCr = $$\frac{n+1}{r+1}=\frac{8}{3}$$
So, 3n + 3 = 8r + 8
or, 3 × 3r + 3 = 8r + 8
or, r = 8 – 3 =5
Hence, n = 3r = 3 × 5 = 15

Question 69.
In how many ways 4 members can occupy 9 vacant seats in a row: [1 Mark, Dec. 2016]
(a) 3204
(b) 3024
(c) 49
(d) 94
(b) is correct
Total ways = 9P4 = 9.8.7.6 = 3024

Question 70.
The number of arrangements that can be formed from the letters of the word “ALLAHABAD”. [1 Mark, June 2017]
(a) 7560
(b) 3780
(c) 30240
(d) 15320
$$\frac{9 !}{(4 !) \cdot(2 !)}=\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5(4 !)}{(4 !) \cdot 2 \cdot 1 .}$$ = 7560
Option (a) is correct

Question 71.
If 10C3 + 2.10C4 + 10C5 = nC5 then the value of n = .[1 Mark, June 2017]
(a) 10
(b) 11
(c) 12
(d) 13
10C3 + 2.10C4 + 10C5 = nC5
= 10C3 + 10C4 + 10C4 + 10C5 = nC5
or; 11C4 + 11C5 = nC5
or; 12C5 = nC5 ⇒ n = 12
Option (c) is correct

Question 72.
The 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]
(a) 360
(b) 90
(c) 180
(d) 45
No. of parallelograms = 6C2 . 4C2
= 15 × 6 = 90
Option (b) is correct

Question 73.
If nP13 : (n+1)P12 = 3 : 4 then ‘n’ is : [1 Mark, Dec. 2017]
(a) 13
(b) 15
(c) 18
(d) 31
(b) is correct
Tricks : Go by choices (Test from beginning)

Question 74.
In 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]
(a) 29,950
(b) 25,940
(c) 25,920
(d) None of these
(c)
From Qts. ; Total no. of ways
= (3!) . (3!) . (5!) . (3!)
= 6 × 6 × 120 × 6
= 25,920.

Question 75.
If 12C3 + 2.12C4 + 12C5 = 12C5,The value of x : [1 Mark, June 2018]
(a) 3 or 5
(b) 5 or 9
(c) 7 or 1
(d) 9 or 12
(b)
12C3 + 2.12C4 + 12C5 = 12C5
12C3 + 12C4 + 12C4 + 12C5 = 14Cx
13C4 + 13C5 = 14Cx
14C5 = 14C14-5 = 14Cx
⇒ x = 14 – 5 = 9
x = 5 or 9

Question 76.
The number of ways in which a man can invite one or more of his 7 friends to dinner is
(a) 64
(b) 128
(c) 127
(d) 63
(c)
No. of ways to invite at least one friend = 27 – 1
= 128 – 1
= 127.

Question 77.
The number of words from the letters of the word BHARAT, in which B and H will never come together, is:[1 Mark, Nov. 2018]
(a) 120
(b) 360
(c) 240
(d) None
(c)
Use Gap rule.
*A * R * A * T *
Total words = $$\frac{4 !}{2 !}$$ × 5P2
= $$\frac{24}{2}$$ × 5 × 4 = 240

Question 78.
The value of N in $$\frac{1}{7 !}+\frac{1}{8 !}=\frac{N}{9 !}$$ is [1 Mark, Nov 2018]
(a) 81
(b) 64
(c) 78
(d) 89
(a)

⇒ N = 81

Question 79.
If nPr = 720 and nCr = 120 then r is: [1 Mark, Nov. 2018]
(a) 4
(b) 5
(c) 3
(d) 6
(c)
nPr = nCr.r!
or; 720 = 120 (r!)
or 6 = r! ⇒ r! = 3!
r = 3

Question 80.
A 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]
(a) 46
(b) 64
(c) 86
(d) None