4
Q:

# A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady?

 A) 123 B) 113 C) 246 D) 945

Explanation:

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.

(i) 1 lady out of 4 and 4 gentlemen out of 6

(ii) 2 ladies out of 4 and 3 gentlemen out of 6

(iii) 3 ladies out of 4 and 2 gentlemen out of 6

(iv) 4 ladies out of 4 and 1 gentlemen out of 6

In case I the number of ways = $\inline 4C_{1}\times 6C_{4}$ = 4 x 15 = 60

In case II the number of ways = $\inline 4C_{2}\times 6C_{3}$ = 6 x 20 = 120

In case III the number of ways = $\inline 4C_{3}\times 6C_{2}$ = 4 x 15 = 60

In case IV the number of ways = $\inline 4C_{4}\times 6C_{1}$ = 1 x 6 = 6

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

Q:

A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group ?

 A) 600 B) 610 C) 609 D) 599

Explanation:

Two men, three women and one child can be selected in ⁴C₂ x ⁶C₃ x ⁵C₁ ways

= $\inline \fn_jvn \small \left ( \frac{4x3}{2x1} \right )x \left ( \frac{6x5x4}{3x2x1} \right )x 5$

= 600 ways.

1 10
Q:

The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ?

 A) 2580 B) 3687 C) 4320 D) 5460

Explanation:

Let 'Y' be the youngest player.

The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.

The required number of ways = 3(3!) = 4320.

3 73
Q:

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most  ?

 A) 215 B) 268 C) 254 D) 216

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 - 1 = 215.

1 35
Q:

The number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is  ?

 A) 2(6!) B) 6! x 7 C) 6! x ⁷P₆ D) None

Answer & Explanation Answer: C) 6! x ⁷P₆

Explanation:

We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.

Hence required number of ways = 6! x ⁷P₆

1 58
Q:

The number of permutations of the letters of the word 'MESMERISE' is  ?

 A) 9!/(2!)^{2}x3! B) 9! x 2! x 3! C) 0 D) None

Number of arrangements = $\inline&space;\fn_jvn&space;\small&space;\frac{9!}{(2!)^{2}x3!}$