Permutation & Combination Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (c)

The committee of 4 persons is to be so formed that it has at least 2 men. The different ways that we can choose to form such a committee are:

(i) 2m. 2w in $^6C_2 × ^4C_2$ = ${6 × 5}/{2 × 1}$ × ${3 × 3}/{2 × 1}$ = 90

(ii) 3m. 1w in $^6C_3 × ^4C_1$ = ${6 × 5 × 4}/{3 × 2 × 1}$ × 4= 80

(iii) 4m in $^6C_4$ = ${6 × 5}/{2 × 1}$ = 15

∴ Total no. of different ways in which a committee of 4 persons can be formed so that it has at least 2 men.

= 90 + 18 + 15 = 185

Answer: (e)

One girl can be chosen in $^4C_1$ = 4 ways

and 4 boys can be chosen in $^6C_4$ = 15 ways

∴ Total number of ways = 4 × 15 = 60 ways

Answer: (a)

In MATHEMATICS, the consonants M and T are repeated two times each.

Also the vowel A is repeated two times.

Since there are four vowels, A, A, E and I; A being repeated,

therefore vowels can be arranged in $4!/2$ = 12 ways.

Now remaining 7 consonants, with M, T being repeated,

can be written in $7!/{2 × 2}$ = 7 × 6 × 5 × 3 × 2 = 1260 ways.

Now four vowels together can take any of the 8 places as shown below:

VC VC VC VC VC VC VC V

∴ Total number of ways in which the letters of the word MATHEMATICS can be arranged such that vowels always

come together = 1260 × 8 × 12 = 120960.

Answer: (b)

The student can answer by using the following combinations 4 from 5 and 6 from 8 or 5 from 5 and 5 from 8

i.e. $^5C_4 × ^8C_6 + ^5C_5 × ^8C_5$ = ${5!}/{4!1!}$ × ${8!}/{6!2!}$ + ${5!}/{5!0!}$ × ${8!}/{3!5!}$

= 140 + 56 = 196 Ways

Answer: (a)

Since 2 players are always included

∴ We have to select only 9 players from 20.

∴ No. of ways = $^{20}C_9$ . Also 4 are always excluded. i.e. 9 should be selected from 16 only, i.e. $^{16}C_9$ ways.