2
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

Answer:   B) 1587600

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:

If (1 × 2 × 3 × 4 ........ × n) = n!, then 15! - 14! - 13! is equal to ___?

 A) 14 × 13 × 13! B) 15 × 14 × 14! C) 14 × 12 × 12! D) 15 × 13 × 13!

Answer & Explanation Answer: D) 15 × 13 × 13!

Explanation:

15! - 14! - 13!

= (15 × 14 × 13!) - (14 × 13!) - (13!)

= 13! (15 × 14 - 14 - 1)

= 13! (15 × 14 - 15)

= 13! x 15 (14 - 1)

= 15 × 13 × 13!

Report Error

4 25
Q:

To fill 8 vacancies there are 15 candidates of which 5 are from ST. If 3 of the vacancies are reserved for ST candidates while the rest are open to all, Find the number of ways in which the selection can be done ?

 A) 7920 B) 74841 C) 14874 D) 10213

Answer & Explanation Answer: A) 7920

Explanation:

ST candidates vacancies can be filled by 5C3 ways = 10

Remaining vacancies are 5 that are to be filled by 12

=> 12C5 = (12x11x10x9x8)/(5x4x3x2x1) = 792

Total number of filling the vacancies = 10 x 792 = 7920

Report Error

9 330
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.

Report Error

5 194
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

Answer & Explanation Answer: A) 43929

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

Report Error

8 325
Q:

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

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

Answer & Explanation Answer: 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.

Report Error

11 548