Mathematical Permutation & Combination MCQ | Multiple Choice Questions -

Last Updated on: 3rd January 2025, 12:05 pm

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Mathematical Permutation & Combination MCQ

1. If ³⁵C _n₊₇= ³⁵C _{4 n}_–₂, find value of n

(a) 20

(b) 10

(d) 6.

³⁵C _n₊₇= ³⁵C ₍_{4 n}_–_{2), or,}n + 7 = 35 – 4n – 2, Or, n + 7 = 35 – 4n + 2, Or, n + 4n = 35 + 2 – 7 = 30

Or, 5n = 30, Or, n = 6. So, option (d) is correct.

2. In how many ways can 10 people line up at a ticket window of a cinema hall?

(a) 3628800

(b) 3580800

(d) 3304400.

10 people can line up at a ticket window of a cinema hall in 10! Ways=

1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3628800 ways. So, option (a) is correct.

3. Ten students are participating in a race. In how many ways can the first three prizes be won?

(a) 900

(b) 750

(d) 720.

3 students should be chosen out of 10 students for 1st three prizes in ¹⁰P₃ ways
$\displaystyle =\frac{{\left( {10!} \right)}}{{\left( {10-3} \right)!}}$ Ways

= 10 x 9 x 8 = 720 ways.

So, option (d) is correct.

4. If C(n, 7) = C(n, 5) find n

(a) 16

(b) 12

(d) 14.

C(n, 7) = ⁿC₇ = n!/ {(7! X (n – 7)!}. Again, C(n, 5)
$\displaystyle {{=}^{n}}{{C}_{5}}=\frac{{n!}}{{{(5!\times (n-5)!}}}$
$\displaystyle So,\frac{{n!}}{{{(7!\times (n-7)!}}}=\frac{{n!}}{{{(5!\times (n-5)!}}}$
or, (7! X (n – 7)! = 5! X (n – 5)!}.

$\displaystyle \frac{{\left( {n\text{ }-\text{ }7} \right)!}}{{\left( {n\text{ }-\text{ }5} \right)!}}=\frac{{7!}}{{5!}}\text{ }Or,\text{ }\frac{{\left( {n\text{ }-\text{ }5} \right)\text{ }\left( {n\text{ }-\text{ }6} \right)\text{ }\left( {n\text{ }-\text{ }7} \right)!\text{ }}}{{\left( {n\text{ }-\text{ }7} \right)!\text{ }}}=\frac{{7\times 6\text{ }\times 5!}}{{5!}}$

Or, (n – 5) (n – 6) = 7 X 6, or (n – 5) (n – 6) = 42, Or, n² – 11x + 30 = 42

Or, n² – 11x – 12 = 0, Or, n² + n – 12x – 12 = 0, Or, n(n + 1) – 12(n + 1) = 0

Or, (n + 1) (n – 12) = 0 \n ¹1 n = 12. So, option (b) is correct.

5. In how many ways can 5 sportsmen be selected from a group of 10

(a) 225

(b) 290

(d) 230.

5 sportsmen can be selected from a group of 10 in ¹⁰C₅ways

$\displaystyle =\frac{{10!}}{{5!\times \left( {10\text{ }-\text{ }5} \right)!}}\text{ }Ways$

$\displaystyle =\text{ }\frac{{10!}}{{\left( {5!\times 5!} \right)}}\text{ }ways\text{ }=\text{ }\frac{{10\text{ }\times \text{ }9\text{ }\times \text{ }8\text{ }7\text{ }\times \text{ }6\text{ }\times \text{ }5!}}{{\left( {5!\text{ }\times \text{ }5!} \right)}}\text{ }ways\text{ }$
$\displaystyle =\text{ }\frac{{10\text{ }\times \text{ }9\text{ }\times \text{ }8\text{ }\times \text{ }7\text{ }\times \text{ }6}}{{1\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }4\text{ }\times \text{ }5}}\text{ }$

= 9 x 4 x 7 = 252 ways. So, option (c) is correct.

6. In how many ways can a cricket team of 11 players be selected of 16 players if one particular player is to be excluded?

(a) 1560

(b) 1365

(d) 1250.

Specific one player must be excluded. So, 11 players should be chosen out of (16 – 1) = 15 players in

$\displaystyle ^{{15}}{{C}_{{11}}}ways=\frac{{15!}}{{(11!\text{ }\times \left( {15\text{ }-\text{ }11} \right)!}}ways\text{ }$
$\displaystyle =\text{ }\frac{{15!}}{{11!\text{ }\times \text{ }4!}}\text{ }ways\text{ }=\text{ }\frac{{\left( {15\text{ }\times \text{ }14\text{ }\times \text{ }13\text{ }\times \text{ }12\text{ }\times \text{ }11!} \right)}}{{11!\text{ }\times \text{ }\left( {1\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }4} \right)}}ways.\text{ }$
= 15 x 7 x 13 = 1365 ways. So, option (b) is correct.

7. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls, if each selection consists of 3 balls of each colour.

(a) 3500

(b) 1500

(d) 6000.

3 red balls will be chosen out of 6 red balls in ⁶C_{3 ways}. 3 white balls will be chosen out of 5 white balls in ⁵C_{3 ways.}3 blue balls will be chosen out of 5 blue balls, ⁵C₃ways.

8 There are eight chairs in a room. In how many ways 5 students can sit on them?

(a) 6100

(b) 6720

(d) 6500

9 There are six vacant seats in a railway compartment. In how many ways 2 persons can sit there?

(a) 30

(b) 25

(d) 45

2 seats should be chosen out of 6 seats. So, numbers of arrangement = ⁶P₂
$\displaystyle =\frac{{6!}}{{\left( {6-2} \right)!}}=\frac{{6!}}{{4!}}=6\times 5\text{ }=30\text{ }ways.\text{ }$
So, option (a) is correct.

10 There are 10 seats in a bus. In how many ways 4 passengers acquire the 10 seats.

(a) 4500

(b) 5090

(d) 5040

Number of ways 4 passenger can sit out of 10 seats = ¹⁰P₄

$\displaystyle =\frac{{10!}}{{\left( {10-4} \right)!\text{ }}}=\frac{{10!}}{{6!}}$

=10 X 9X 8X 7= 5040.

So, option (d) is correct.

11. How many permutations will be from the letters of word ECONOMICS ?

(a) 89,000

(b) 90,420

(d) 90,100

The word “ECONOMICS” is composed of 9 letters out of which O is repeated twice, C is repeated twice. So, Number of ways the letters can be arranged

$\displaystyle =\frac{{9!}}{{\left( {2!\text{ }\times 2!} \right)}}=\frac{{\left( {9\times 8\times 7\times 6\times 4\times 5\times 3\times 2} \right)}}{4}\text{ }$

= 9X8X7X6X5X3X2 = 90720. So, option (c) is correct.

12. If ⁿP₄ = 12 x ⁿP₂_¢ find value of n.

(a) 2

(b) 6

(d) 3

ⁿP₄= n! / {(n-4) ! X 4!}, ⁿP₂= n! / {(n-2) ! X 2 !}.

$\displaystyle So,\text{ }~\left[ {\frac{{n!}}{{{\left( {n-4} \right)!\text{ }\times \text{ }4!}}}} \right]=\text{ }12\times \left[ {\frac{{n!}}{{{\left( {n-2} \right)!\text{ }\times \text{ }2!}}}} \right].$

$\displaystyle So,\text{ }\frac{1}{{\left( {n-4} \right)!}}=\frac{{12}}{{\left( {n-2} \right)!}}\text{ }\left[ {dividing\text{ }by\text{ }n!\text{ }both\text{ }sides} \right].\text{ }$

$\displaystyle Or,\text{ }\frac{{4!}}{{\left( {n-4} \right)!}}=\frac{{12}}{{\left( {n-2} \right)!}}$

$\displaystyle Or,\text{ }\frac{{4!}}{{\left( {n-4} \right)!}}=\frac{{12}}{{\left( {n-2} \right)\text{ }\times \text{ }\left( {n-3} \right)\text{ }\times \text{ }\left( {n-4} \right)!}}$

Or, (n – 2) (n – 3) = 12, Or, n² – 5n + 6 = 12

Or, n² – 5n – 6 = 0, Or, n² + n – 6n – 6 = 0, Or, n (n + 1) – 6 (n – 6) = 0, n = 6 (as n ¹ -1 , because in factorial, all numbers are positive integers). So, option (b) is correct.

13. If ⁿP₃ : ⁿP₂ = 3 : 1, find n.

(a) 8

(b) 6

(d) 3

14. The number of arrangements of the letters in the word FAILURE, so that vowels are always coming together is

(a) 576

(b) 590

(d) 610

The word FAILURE is composed of 7 letters out of which 4 are vowels (A, I U,E). So, considering 4 vowels as one letter (as all 4 vowels will always come together), (F, L, R (A, I, U, E), we have only 4 letters to arrange. . So, number of arrangement keeping vowels together = 4 ! ways = 4 x 3 x 2 x 1 = 24 ways.

Again the vowels can be arranged among themselves in 4 ! ways = 4 x 3 x 2 x 1 = 24 ways

So, total number of arrangement = 24 x 24 = 576. So, option (a) is correct.

15. The number of ways the letters of the word “Triangle” to be arranged so that the word ‘angle’ will be always present is

(a) 30

(b) 50

(d) 18

In the arrangement, the word “angle” will be always present. Considering it as one letter, there are 4 letters “TRI (ANGLE) in the word TRIANGLE. So, Number of ways letter can be arranged =

4 ! ways = 4 x 3 x 2 x 1 = 24 ways. So, option (c) is correct.

16. If ¹⁸C_r = ¹⁸C_r₊₂_¢ the value ^rC₅is

(a) 51

(b) 60

(d) 59

17. If ⁿC₁₀ = ⁿC₁₄_¢ then ²⁵C_nis

(a) 20

(b) 25

(d) 18

ⁿC₁₀ = ⁿC_14..Now ⁿC₁₄= ⁿC_(n_–₁₄), So,ⁿC₁₀ = ⁿC_n_–_14,So, 10 = n – 14, Or, n = 24

So, ²⁵C_n = ²⁵C₂₄= ²⁵C₂₅_–₂₄= ²⁵C₁ = 25. So, option (b) is correct.

18. You are selecting a cricket team of first 11 players out of 16 including 4 bowlers and 2 wicket-keepers. In how many ways you can do it so that the team contains 3 bowlers and 1 wicket-keeper?

(a) 960

(b) 800

(d) 457

Total players = 16, consisting of 4 Bowlers, 2 wicketkeepers. So 16- (4+2) = 10 Batsmen.

So, Number of ways a team of 11 players including 7 Batsmen out of 10, 3 Bowlers out of 4, & 1 wicket keeper out of 2, can be selected = ¹⁰C₇ x ⁴C₃x ²C₁

19. In a Club of 55 members. Every member posts a greeting card to all other members. How many greeting cards were posted by them?
a) 54X55
b) ⁵⁴P₅₅
c) ⁵⁵P₅₄
d) None of the above
One member can post greeting cards to the remaining 54 players in 54 ways. This way, rest of the members can send 54 card. So, The total numbers of greeting cards posted are 54 + 54 + 54 … 55 times= 54 x 55 = 2970
So option (a) is correct

20. Value of ⁿP_n-1 is

We know, ⁿP_r = $\displaystyle \frac{{n!}}{{\left( {n-r} \right)!}}$

So, ⁿP_n-1 = $\displaystyle \left( {\frac{{n!}}{{\left[ {n-(n-1)} \right]!}}} \right)=\frac{{n!}}{{1!}}$ = n!

So option (b) is correct

21. If ⁿP₅ = 60 (ⁿ⁻¹P₃), the value of n is

ⁿP₅ = 60(ⁿ⁻¹P₃)
So, $\displaystyle \left( {\frac{{n!}}{{(n-5)!}}} \right)=60\left( {\frac{{(n-1)!}}{{\left[ {(n-1)-3} \right]!}}} \right)$
On solving the above equation, we get n= -6 and n=10.
But n cannot be negative. So, n=10
Students should solve the equation and show the working
So option (b) is correct

22. In how many ways 8 distinct toys can be distributed among 5 children?

⁸P₅
⁵P₈
5⁸
8⁵

Number of toys = 8, Number of children = 5.
Note that all the toys are distinct. Hence, the number of ways 8 distinct toys can be distributed among 5 children is 5×5×5×5×5×5×5×5=5⁸.
So option (c) is correct

23. In how many ways can we paint the six faces of a cube dice with six different colours?

30
6
6!
None of the above

Let us mark the colours of each face as Col-1……Col-6
Ler us assume that facing up is col-1. So, the face of the cube at the bottom can be painted in 5 different ways. And 4 faces on the horizontal side of the cube are in circular permutation and they can be painted in (4-1)! ways.
Hence, the total number of ways we can paint the faces of a cube with six different colours is 5×(4-1)! Ways = 5X3! = 5X6=30 ways
So option (a) is correct

24. Joining 20 points in a plane, with if 5 points, colinear, how many triangles can be formed?
a) 1130
b) 550
c) 1150
d) 1140
Number of points n = 20. colinear points m = 5. A Triangle cannot be formed by joining collinear points.
Number of triangles from by joining n points of which m are colinear = ⁿC₃ – ^mC₃
So, Number of triangles = ²⁰C₃ – ⁵C₃ = 1140-10 = 1130
So option (a) is correct

25. In how many ways 30 people in a group can shake hands with each other
a. 870
b. 435
c. 900
d. None of the above
There are 30 people. a handshake can be done by 2 people. So, the number of handshakes would be ³⁰C₂ = $\displaystyle \left( {\frac{{30!}}{{2!(30-2)!}}} \right)=\frac{{30\times 29}}{2}$ = 435
So option (b) is correct

26. In a lawn, there are 9 chairs (2 wooden, 3 Plastic, and 4 Steel). In how many ways you can choose 3 chairs so that at least one plastic chair is included?
a. 32
b. 30
c. 64
d. 84
The simplified ways is to find total possible ways of selection – no plastic chair selection.
Total ways of selection of 3 chairs from 9 chairs = ⁹C₃ = 84 ways.
Total number of non plastic chairs = 2+4=6.
3 chairs may be selected from 6 non plastic chairs in ⁶C₃ways = 20 ways
So, 3 chairs with at least one plastic chair can be selcted in 84-20=64 ways
So option (c) is correct

27. In how many ways, a committee of 5 persons, with at least 3 men are to be created can be created by selection from a group of 7 men and 6 women,
(a) 564
(b) 645
(c) 735
(d) 756
The possible ways of having at least 3 men is
(i) 3 men + 2 women, (ii) 4 men + 1 woman, (iii) 5 men,
So, the possible ways are
(⁷C₃ x ⁶C₂) + (⁷C₄ x ⁶C₁) + (⁷C₅) = (525 + 210 + 21)=756
So option (d) is correct

Mathematical Permutation & Combination MCQ | Multiple Choice Questions

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