First Approach:

A number is divisible by 66 if it is divisible by 22 and 33.

Now the possible combinations I found are

(1,3,2)(1,3,2)

(3,1,2)(3,1,2)

(2,3,4)(2,3,4)

(3,2,4)(3,2,4)

(4,3,2)(4,3,2)

(3,4,2)(3,4,2)

(3,5,4)(3,5,4)

(5,3,4)(5,3,4)

total 88 ways.

Second Approach

Case1: unit digit can be filled in only two ways (2,4)(2,4) for nos (3,2,4)(3,2,4)

Tens digit can be filled in 22 ways

Hundred digit can be filled in 11 ways

the required number is 2*1*2=4 ways

Case2: unit digit can be filled in only one ways (2)(2) for nos (1,2,3)(1,2,3)

Tens digit can be filled in 11 ways

Hundred digit can be filled in 22 ways

the required number is 2*1*1=2 ways

Case3: unit digit can be filled in only one ways (4)(4) for nos (3,4,5)(3,4,5)

Tens digit can be filled in 11 ways

Hundred digit can be filled in 22 ways

the required number is 2*1*1=2 ways

So, Total ways=8

DIRECTOR

Total Letters = 8

Vowels=3

Consonants= 5

Considering all vowels together as 1 group and all consonants together as another group.

We have 2 groups one of vowels and other of consonants.

So, Total permutation of 2 groups = 2!=2 ways.

Considering group of vowels permutations possible in the group of vowels =3!=6 ways.

Considering group of consonants permutations possible in the group of consonants =5!/(2!)=5×4×3=60 ways

Hence ,Total permutations =2×6×60=720

In the word ‘MATHEMATICS’, we’ll consider all the vowels AEAI together as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice
Number of ways of arranging these letters =8! / ((2!)(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

The word ‘OPTICAL’ contains 77 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 55 letters can be arranged in 5!=1205!=120ways. The vowels (OIA) can be arranged among themselves in 3!=63!=6 ways. Required number of ways =(120∗6)=720=(120∗6)=720.

The number of arrangement in which A and B are not together

Task 2 must be assigned to P₄ or P₃ =2C₁

For task 1 only 2 persons i.e.., P₃ or P₄and P₅ = 2C₁

Task 3 only 3 person available =3C₁

For task 4 only 2 persons available =2C₁

For task 5 only 1 person left = 1

Therefore, total number of ways

= 2C₁ x 2C₁x 3C₁x 2C₁ x1 = 2x2x3x2x1 =24

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).

As there are six players, So total ways in which they can be arranged = 6!ways =720.

A number of ways in which Ajeet and Mukherjee are together = 5!x2 = 240.

Therefore, Number of ways when they don’t remain together = 720 -240 =480.

The number of ways of selecting three men, two women and three children is:

The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.

The given digits are six.

The number of ways to select two men and three women = ⁵C₂ * ⁶C₃

Total number of letters = 8

Case 1: Exactly one box contains a blue ball

One blue ball can be placed into any of the 6 boxes. i.e, 6 ways of doing this.

Red balls can be filled in the remaining boxes. Since red balls are identical, there is only 1 way of doing this.

Total number of ways $=6\times 1=6$

Case 2: Exactly two boxes contain blue balls

Two blue balls can be placed into (box 1 and box 2) or (box 2 and box 3) or (box 3 and box 4) or (box 4 and box 5) or (box 5 and box 6). i.e, 5 ways of doing this.

Red balls can be filled in the remaining boxes. Since red balls are identical, there is only 1 way of doing this.

Total number of ways $=5\times 1=5$

Case 3: Exactly three boxes contain blue balls

Three blue balls can be placed into (box 1 , box 2 and box 3) or (box 2, box 3 and box 4) or (box 3 , box 4 and box 5) or (box 4, box 5 and box 6). i.e, 4 ways of doing this.

Red balls can be filled in the remaining boxes. Since red balls are identical, there is only 1 way of doing this.

Total number of ways $=4\times 1=4$

Case 4: Exactly four boxes contain blue balls

Four blue balls can be placed into (box 1 , box 2, box 3 and box 4) or (box 2, box 3, box 4 and box 5) or (box 3 , box 4, box 5 and box 6). i.e, 3 ways of doing this.

Red balls can be filled in the remaining boxes. Since red balls are identical, there is only 1 way of doing this.

Total number of ways $=3\times 1=3$

Case 5: Exactly five boxes contain blue balls

Five blue balls can be placed into (box 1, box 2, box 3, box 4 and box 5) or (box 2, box 3, box 4, box 5 and box 6). i.e, 2 ways of doing this.

Red balls can be filled in the remaining boxes. Since red balls are identical, there is only 1 way of doing this.

Total number of ways $=2\times 1=2$

Case 6: All the six boxes contain blue balls

Six blue balls can be placed into (box 1 , box 2, box 3, box 4, box 5 and box 6). i.e, only 1 way of doing this.

Total number of ways = 1

Hence, required number of ways
$=6+5+4+3+2+1=21$