Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.

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: 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: 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: 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: 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: 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

Correct Answer: 7200

Explanation:

From 5 consonants, 3 consonants can be selected in  5 C 3 ways.

 

From 4 vowels, 2 vowels can be selected in  4 C 2 ways.

 

Now with every selection, number of ways of arranging 5 letters is 5 P 5ways.

 

Total number of words =  5 C 3 * 4 C 2 * 5 P 5

 

                                = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

Correct Answer: 105

Explanation:

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room, 

 Then, ⁷C₁ x ⁶C₂ x ⁴C₄ 

_{= 7 x 15 x 1 = 105}

Correct Answer: 120

Explanation:

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

 

There are 6 letters in thegiven word.

 

First arrange 3 vowels.

 

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

 

Remaining 3 letters can be placed in 3 places = 3! ways

 

Total number of possible ways of arranging letters of OLIVER = 3! x  C 3 6 ways = 6x5x4 = 120 ways.

Correct Answer: 43929

Explanation:

Number of words with 5 letters from given 9 alphabets formed =  9 5

 

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is =  9 P 5

 

Number of words can be formed which have at least one letter repeated =   9 5 - 9 P 5

 

= 59049 - 15120

 

= 43929