How many words can be formed from the letters of the word "AFTER", so that the vowels never comes together.

Answer: Option C

Explanation:

Friends the main point to note in this question is letter "P" is written twice in the word.
Easy way to solve this type of permutation question is as,

So word APPLE contains 1A, 2P, 1L and 1E
Required number =
\begin{aligned}
= \frac{5!}{1!*2!*1!*1!} \\
= \frac{5*4*3*2!}{2!} \\
= 60
\end{aligned}

Answer: Option B

Explanation:

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

Answer: Option D

Explanation:

In above word, there are 2 "R" and 2 "U"
So Required number will be

\begin{aligned}
= \frac{6!}{2!*2!} \\
= \frac{6*5*4*3*2*1}{4} \\
= 180
\end{aligned}

Answer: Option B

Explanation:

This question seems to be a bit typical, isn't, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways

Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24

Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.