If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM ?

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: 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: 5040

Explanation:

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

Correct Answer: 17280

Explanation:

Let 4 girls be one unit and now there are 6 units in all.

 

They can be arranged in 6! ways.

 

In each of these arrangements 4 girls can be arranged in 4! ways. 

 

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

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: 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.

Correct Answer: 5^8

Explanation:

As the toys are distinct and not identical,

For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are  5 8 ways to distribute the toys.

 

Hence, it is 5 8 and not 8 5.

Correct Answer: 1440

Explanation:

Given word is THERAPY.

Number of letters in the given word = 7

Number of vowels in the given word = 2 = A & E

Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is

6! x 2! = 720 x 2 = 1440.

Correct Answer: 330

Explanation:

Here the order of choosing the elements doesn?t matter and this is a problem in combinations.

 

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

 

This can be done in 11 C 4 ways = 330 ways

Correct Answer: 576

Explanation:

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.

 

There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4 P 3 ways and 4 constants can be arranged in  4 P 4 ways.

 

Number of words = 4 P 3  x  4 P 4= 24 x 24 = 576