Competitive Exams, Entrance Exams are conducted on MCQ. **Click HERE** to understand the Forms, Structure, Rules of MCQ, techniques of understanding, analysing and selection correct answer of MCQ,**Play the Video** explaining some interesting aspects of selection of correct answer of MCQ.**Complete resources on Mathematics** **Complete resources on Permutation & Combination**

**Permutation & Combination MCQ**

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

(a) 20

(b) 10

(c) 4

(d) 6.

^{35}C _{n }_{+}_{ 7 }= ^{35}C _{(}_{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

(c) 324600

(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

(c) 950

(d) 720.

3 students should be chosen out of 10 students for 1st three prizes in ^{10}P_{3} 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

(c) 20

(d) 14.

C(n, 7) = ^{n}C_{7} = n!/ {(7! X (n – 7)!}. Again, C(n, 5)

or, (7! X (n – 7)! = 5! X (n – 5)!}.

Or, (n – 5) (n – 6) = 7 X 6, or (n – 5) (n – 6) = 42, Or, n^{2} – 11x + 30 = 42

Or, n^{2} – 11x – 12 = 0, Or, n^{2} + 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

(c) 252

(d) 230.

5 sportsmen can be selected from a group of 10 in ^{10}C_{5 }ways

= 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

(c) 1150

(d) 1250.

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

= 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

(c) 2000

(d) 6000.

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

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

(a) 6100

(b) 6720

(c) 5500

(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

(c) 75

(d) 45

2 seats should be chosen out of 6 seats. So, numbers of arrangement = ^{6}P_{2 }

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

(c) 4990

(d) 5040

Number of ways 4 passenger can sit out of 10 seats = ^{10}P_{4 }

=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

(c) 90,720

(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

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

**12. If **^{n}**P**_{4}** = 12 x **^{n}**P**_{2}_{¢}** find value of n.**

(a) 2

(b) 6

(c) 8

(d) 3

Or, (n – 2) (n – 3) = 12, Or, n^{2} – 5n + 6 = 12

Or, n^{2} – 5n – 6 = 0, Or, n^{2} + 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 **^{n}**P**_{3}** : **^{n}**P**_{2}** = 3 : 1, find n.**

(a) 8

(b) 6

(c) 5

(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

(c) 490

(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

(c) 24

(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 **^{18}**C**_{r }** = **^{18}**C**_{r }_{+}_{ 2}_{¢}_{ }** the value **^{r}**C**_{5 }**is**

(a) 51

(b) 60

(c) 56

(d) 59

**17. If **^{n}**C**_{10 }** = **^{n}**C**_{14}_{¢}_{ }** then **^{25}**C**_{n }**is**

(a) 20

(b) 25

(c) 27

(d) 18

^{n}C_{10 } = ^{n}C_{14.. }Now ^{n}C_{14 }= ^{n}C_{(n }_{–}_{ 14}), So,_{ }^{n}C_{10 } = ^{n}C_{n }_{–}_{ 14, }So, 10 = n – 14, Or, n = 24

So, ^{25}C_{n} = ^{25}C_{24 }= ^{25}C_{25 }_{–}_{ 24}^{ }= ^{25}C_{1} = 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

(c) 650

(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 = ^{10}C_{7} x ^{4}C_{3 }x ^{2}C_{1}