In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together ?

Correct Answer: n = 29/5

Explanation:

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1) 

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

 

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5

Correct Answer: 1170

Explanation:

Possibilities     Bowlers      Batsmen         Number of ways

 

                         6               9

 

         1              4                7               ( 6 C 4 * 9 C 7 )

 

         2              5                6               6 C 5 * 9 C 6

 

         3              6                5               6 C 6 * 9 C 5

 

6 C 4 * 9 C 7 = 15 x 36 = 540

 

6 C 5 * 9 C 6 = 6 x 84 = 504

 

  6 C 6 * 9 C 5= 1 x 126 = 126

 

Total = 1170

Correct Answer: 35

Explanation:

A triangle is formed by joining any three non-collinear points in pairs.

 

There are 7 non-collinear points

 

The number of triangles formed =  7 C 3 = 35

Correct Answer: 59

Explanation:

The first letter is E and the last one is R.

 

Therefore, one has to find two more letters from the remaining 11 letters.

 

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

 

The second and third positions can either have two different letters or have both the letters to be the same.

 

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

 

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

 

Total number of possibilities = 56 + 3 = 59

Correct Answer: 191

Explanation:

Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind.

Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191

Correct Answer: 5040

Explanation:

The 7 letters word 'POVERTY' be arranged in  7 P 7 ways = 7! = 5040 ways.

Correct Answer: 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.

Correct Answer: 3600

Explanation:

Let the beds be numbered 1 to 7.

 

Case 1 : Suppose Anju is allotted bed number 1. 

Then, Parvin cannot be allotted bed number 2. 

So Parvin can be allotted a bed in 5 ways. 

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

 

Case 2 : Anju is allotted bed number 7. 

Then, Parvin cannot be allotted bed number 6 

As in Case 1, the beds can be allotted in 600 ways.

 

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. 

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju?s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. 

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

Correct Answer: 32

Explanation:

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

Correct Answer: 720

Explanation:

Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240

 

Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.