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 ?

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: 147320

Explanation:

Sum of 4 digit numbers = (2+4+6+8) x  P 3 3 x (1111) = 20 x 6 x 1111 = 133320 

Sum of 3 digit numbers = (2+4+6+8) x  P 2 3 x (111) = 20 x 6 x 111 = 13320 

Sum of 2 digit numbers = (2+4+6+8) x  P 1 3 x (11) = 20 x 3 x 11 = 660 

Sum of 1 digit numbers = (2+4+6+8) x  P 0 3 x (1) = 20 x 1 x 1 = 20 

 

Adding All , Sum = 147320

Correct Answer: 23

Explanation:

As in this problem , buying any fruit is different case , as buying apple is independent from buying banana. so ADDITION rule will be used.

 

C 1 9 + C 1 8 + C 1 6 = 23 will be answer.

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: 60

Explanation:

Here given the required digit number is 4 digit.

It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available.

x x x 5

The remaining places can be filled by remaining digits as 5 x 4 x 3 ways.

 

Hence, number 4-digit numbers can be formed are 5 x 4 x 3 = 20 x 3 = 60.

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