8
Q:

# How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?

 A) 216 B) 45360 C) 1260 D) 43200

Explanation:

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

$\inline \therefore$ The number of ways in which 9 letters can be arranged = $\inline \frac{9!}{2!\times 2!\times 2!}$ = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in $\inline \frac{6!}{2!\times 2!}$ = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in $\inline \frac{4!}{2!}$ = 12 ways.

$\inline \therefore$ The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

Q:

A,B,C,D,E,F,G and H are sitting around a circular table facing the centre but not necessarily in the same order. G sits third to the right of C. E is second to the right of G and 4th to the right of H. B is fourth to the right of C. D is not an immediate neighbour of E. A and C are immediate neighbours.

Which of the following is/are correct ?

 A) F is third to the left of B B) F is second to the right of B C) B is an immediate neighbour of D D) All of the above

Answer & Explanation Answer: B) F is second to the right of B

Explanation:

From the given information, the circular arrangement is

Here F is second to the right of B and the remaning all are wrong.

3 17
Q:

Nine different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, how many such words can be formed which have at least one letter repeated ?

 A) 43929 B) 59049 C) 15120 D) 0

Explanation:

Number of words with 5 letters from given 9 alphabets formed = $\inline \fn_jvn 9^{5}$

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is = $\fn_jvn 9_{}^{P}\textrm{5}$

Number of words can be formed which have at least one letter repeated = $\inline \fn_jvn 9^{5}$ - $\fn_jvn 9_{}^{P}\textrm{5}$

= 59049 - 15120

= 43929

6 89
Q:

In how many different ways can the letters of the word 'ABYSMAL' be arranged ?

 A) 5040 B) 3650 C) 4150 D) 2520

Explanation:

Total number of letters in the word ABYSMAL are 7

Number of ways these 7 letters can be arranged are 7! ways

But the letter is repeated and this can be arranged in 2! ways

=> Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

8 181
Q:

In how many ways the letters of the word OLIVER be arranged so that the vowels in the word always occur in the dictionary order as we move from left to right ?

 A) 186 B) 144 C) 136 D) 120

Explanation:

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

There are 6 letters in thegiven word.

First arrange 3 vowels.

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

Remaining 3 letters can be placed in 3 places = 3! ways

Total number of possible ways of arranging letters of OLIVER = 3! x 6C3 ways

= 6x5x4 = 120 ways.

7 250
Q:

Find the total numbers greater than 4000 that can be formed with digits 2, 3, 4, 5, 6 no digit being repeated in any number ?

 A) 120 B) 256 C) 192 D) 244

Explanation:

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

The number can be 4 digited but greater than 4000 or 5 digited.

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x 4P3 = 3 x 24 = 72.

5 digited numbers = 5P5 = 5! = 120

So the total numbers greater than 4000 = 72 + 120 = 192