Answer: B

Answer with the explanation:

We can combine 3 consonants and 2 vowels out of 7 consonants and 4 vowels in a way

⁷C₃ * ⁴C₂ = = 210

That means 210 groups having 3 consonants and 2 vowels.

Each group consists of 5 words that mean the possible arrangement of the letters is 5!

Or, 5! = 5*4*3*2*1 = 120 ways.

Therefore, the required number of words = 210*120 = 25200.

Answer: D

Answer with the explanation:

A number is divisible by 5 if the number ends with 0 or 5, but we don't have 0 in the given digits that means 5 should come at the unit place.

Now, one of the remaining 5 digits (2, 3, 4, 6, and 7) can come at the tens place.
Similarly, we can fill the hundreds place by one of the remaining 4 digits.
Therefore, the thousands place can be filled by one of the remaining 3 digits.
Hence, the required number of the numbers = 1*5*4*3 = 60.

Answer: A

Answer with the explanation:

The word CRICKET has 7 different letters, but ATQ, the vowels should always come together.
Now, let the vowels IE as a single entity.
Therefore, the number of letters is CRCKT = 5 in which C is repeated twice, and IE = 1
Since the total number of letters = 5+1 = 6
So the arrangement (permutation) would be = = = 5*4*3= 60 ways.

Note: Here 2! is taken in the denominator, because the letter C is repeated twice.

Now, the vowels IE can be arranged in 2 different ways, i.e., ²P₂ = 2! = 2*1 = 2 ways.

So, the new words that can be formed = 60*2 = 120.

Answer: C

Answer with the explanation:

The word machine consists of 7 letters in which there are 3 vowels and 4 consonants.
ATQ, the vowels A, I, E can be placed at any of the position out of 1, 3, 5, and 7.
That means the number of ways to arrange the vowels = ⁴P₃ = = = 4*3*2*1 = 24

Similarly, the number of ways to arrange the consonants = ⁴P₄ = = 4*3*2*1 = 24

Note: we know that 0! = 1

Therefore the required numbers of ways = 24*24 = 576

Answer: B

Answer with the explanation:

The word ACTORS consist of 6 letters in which there are 2 vowels and 4 consonants
ATQ, the vowels A, O can be placed at any of the position out of 2, 4, and 6.
That means the number of ways to arrange the vowels = ³P₂ = = = 3*2 = 6 ways

Similarly, the number of ways to arrange the consonants = ⁴P₄ = = 4*3*2*1 = 24 ways

Note: we know that 0! = 1

Therefore the required numbers of ways = 6*24 = 144