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Q:

# How many different words can be made using the letters of the word ' HALLUCINATION ' if all constants are together?

 A) 129780 B) 1587600 C) 35600 D) None of these

Explanation:

H   L   C   N    T    A   U   I   O

L        N          A        I

There are total 131 letters out of which 7 are consonants and 6 are vowels. Also ther are 2L's , 2N's, 2A's and 2I's.

If all the consonants  are together then the numberof arrangements = $\inline&space;\frac{7!}{2!2!}$ . But the 7 consonants  can be arranged themselves in $\inline&space;\frac{7!}{2!2!}$ ways.

Hence the required number of ways = $\inline&space;\frac{7!}{2!\times&space;2!}\times&space;\frac{7!}{2!\times&space;2!}$ = $\inline&space;(1260)^{2}$ = 1587600

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 57
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.

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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 41
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

Explanation:

n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.
The letter pattern 'MESMERISE' consists of 10 letters of which there are 2M's, 3E's, 2S's and 1I and 1R.
Number of arrangements = $\inline&space;\fn_jvn&space;\small&space;\frac{9!}{(2!)^{2}x3!}$

0 91
Q:

A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior ?

 A) ²²C₁₀ + 1 B) ²²C₉ + ¹⁰C₁ C) ²²C₁₀ D) ²²C₁₀ - 1

Answer & Explanation Answer: D) ²²C₁₀ - 1

Explanation:

The total number of ways of forming the group of ten representatives is ²²C₁₀.
The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way
The required number of ways = ²²C₁₀ - 1