4
Q:

In the below word how many words are there in which R and W are at the end positions?

RAINBOW

A) 120 B) 180
C) 210 D) 240

Answer:   D) 240



Explanation:

When R and W are the first and last letters of all the words then we can arrange them in 5!ways. Similarly When W and R are the first and last letters of the words then the remaining letters can be arrange in 5! ways.

Thus the total number of permutations = 2 x 5!  = 2 x 120 = 240

Q:

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together

A) 50000 B) 40500
C) 5040 D) 50400
 
Answer & Explanation Answer: D) 50400

Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

 

Thus, we have CRPRTN (OOAIO).

 

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

 

Number of ways arranging these letters =7!/2!= 2520.

 

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 3!/5!= 20 ways.

 

Required number of ways = (2520 x 20) = 50400.

Report Error

View Answer Report Error Discuss

1 12424
Q:

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A) 720 B) 520
C) 700 D) 750
 
Answer & Explanation Answer: A) 720

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Report Error

View Answer Report Error Discuss

84 66411
Q:

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A) 564 B) 735
C) 756 D) 657
 
Answer & Explanation Answer: C) 756

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only). 

 

Required number of ways= 7C3*6C2+7C4*6C1+7C5 = 756.

Report Error

View Answer Report Error Discuss

11 17444
Q:

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

A) 1260 B) 210
C) 210 x 6! D) 1512
 
Answer & Explanation Answer: A) 1260

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

Report Error

View Answer Report Error Discuss

0 4652