In how many ways can the letters of the word "CORPORATION" be arranged so that vowels always come together.

Answer: Option A

Explanation:

word SIGNATURE contains total 9 letters.
There are four vowels in this word, I, A, U and E

Make it as, SGNTR(IAUE), consider all vowels as 1 letter for now
So total letter are 6.
6 letters can be arranged in 6! ways = 720 ways
Vowels can be arranged in themselves in 4! ways = 24 ways

Required number of ways = 720*24 = 17280

Answer: Option B

Explanation:

First thing to understand in this question is that it is a permutation question.
Total number of words = 5
Required number =

$$\begin{aligned} ^5{P}_5 = 5! \\ = 5*4*3*2*1 = 120 \end{aligned}$$

Answer: Option B

Explanation:

$$\begin{aligned} = \frac{30!}{28!} \\ = \frac{30 * 29 * 28!}{28!} \\ = 30 * 29 = 870 \end{aligned}$$

Answer: Option B

Explanation:

As we can see the letter "E" is twice in given word, so Required Number

$$\begin{aligned} = \frac{7!}{2!} \\ = \frac{7*6*5*4*3*2!}{2!} \\ = 2520 \end{aligned}$$

Answer: Option D

Explanation:

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
So we can have
(5 men) or (4 men and 1 woman) or (3 men and 2 woman)

$$\begin{aligned} (^{5}{C}_{5}) + (^{5}{C}_{4} * ^{6}{C}_{1}) + \\ + (^{5}{C}_{3} * ^{6}{C}_{2}) \\ = \left[\dfrac{7 \times 6 }{2 \times 1}\right] + \left[\left( \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right) \times 6 \right] + \\ \left[\left( \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right) \times \left( \dfrac{6 \times 5}{2 \times 1} \right) \right] \\ = 21 + 210 + 525 = 756 \end{aligned}$$