Permutation & Combination MCQ -

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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
$\displaystyle =\frac{{\left( {10!} \right)}}{{\left( {10-3} \right)!}}\text{ Ways}$
= 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

$\displaystyle ^{n}{{P}<em>{4}}=\frac{{n!}}{{\left{ {\left( {n-4} \right)!\text{ }\times \text{ }4!} \right}}}{{,}^{n}}{{P}</em>{2}}=\frac{{n!}}{{\left{ {\left( {n-2} \right)!\text{ }\times \text{ }2!} \right}}}.$

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

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

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

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

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

Permutation & Combination MCQ

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