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Q:

# In a G - 20 meeting there were total 20 people representing their own country. All the representative sat around a circular table. Find the number of ways in which we can arrange them around a circular table so that there is exactly one person between two representatives namely Manmohan and Musharraf.

 A) 2 x (17!) B) 2 x (18!) C) (3!) x (18!) D) (17!)

Explanation:

A person can be choosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf , Manmohan and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways.

$\inline \therefore$ Required number of permutations = 18 x (17!) x 2 = 2 x 18!

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.

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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.

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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

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 55
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 = $\inline&space;\fn_jvn&space;\small&space;\frac{9!}{(2!)^{2}x3!}$