How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Correct Answer: 5040

Explanation:

Given word is GLACIOUS has 8 letters.

=> C is fixed in one of the 8 places

Then, the remaining 7 letters can be arranged in 7! ways = 5040.

Correct Answer: 600

Explanation:

Two men, three women and one child can be selected in ?C? x ?C? x ?C? ways = 600 ways.

Correct Answer: 9!/(2!)^{2}x3!

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 =  9 ! ( 2 ! ) 2 × 3 !

Correct Answer: ²²C?? - 1

Explanation:

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

Correct Answer: C

Explanation:

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

 

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

 

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

 

Hence, number of ways in which we can select the black balls

 

= 4C1 + 4C2 + 4C3 + 4C4
= 2 4 - 1 ........(A)

 

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

 

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

 

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
= 2 3 - 1........(B)

 

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

 

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + ? + 5C5
= 2 5..............(C)

 

From (A), (B) and (C), required number of ways
=   2 5 2 4 - 1 2 3 - 1

Correct Answer: 1/5525

Explanation:

n(S) = 52C3 = 132600/6 = 22100

n(E) = 4C3 = 24/6 = 4

Correct Answer: 2 x (18!)

Explanation:

A person can be chosen 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. 

 

Required number of permutations = 18 x (17!) x 2 = 2 x 18!

Correct Answer: 72

Explanation:

 

The bus fromA to B can be selected in 3 ways.

The bus from B to C can be selected in 4 ways.

The bus from C toD can be selected in 2 ways.

The bus fromD to E can be selected in 3 ways.

So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72

 

Correct Answer: 11760

Explanation:

Required number of ways=  8 C 5 × 10 C 6 = ( 8 C 3 × 10 C 4)= 11760.

Correct Answer: 120960

Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

 

Thus, we have MTHMTCS (AEAI).

 

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

 

Number of ways of arranging these letters = 8!/(2! x 2!)= 10080.

 

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

 

Number of ways of arranging these letters =4!/2!= 12.

 

Required number of words = (10080 x 12) = 120960