From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done

Answer: Option C

Explanation:

We need to find the ways that vowels NEVER come together.
Vowels are A, E
Let the word be FTR(AE) having 4 words.
Total ways = 4! = 24
Vowels can have total ways 2! = 2
Number of ways having vowel together = 48
Total number of words using all letter = 5! = 120

Number of words having vowels never together = 120-48
= 72

Answer: Option D

Explanation:

In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.
So we can have

(four boys) or (three boys and one girl) or (two boys and two girls) or (one boy and three gils)

This combination question can be solved as

\begin{aligned}
(^{6}{C}_{4}) + (^{6}{C}_{3} * ^{4}{C}_{1}) + \\
+ (^{6}{C}_{2} * ^{4}{C}_{2}) + (^{6}{C}_{1} * ^{4}{C}_{3}) \\

= \left[\dfrac{6 \times 5 }{2 \times 1}\right] + \left[\left(\dfrac{6 \times 5 \times 4 }{3 \times 2 \times 1}\right) \times 4\right] + \\\left[\left(\dfrac{6 \times 5 }{2 \times 1}\right)\left(\dfrac{4 \times 3 }{2 \times 1}\right)\right] + \left[6 \times 4 \right] \\
= 15 + 80 + 90 + 24\\
= 209
\end{aligned}

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 D

Explanation:

\begin{aligned}
^{n}{C}_{n} = 1 \\
^{100}{C}_{100} = 1
\end{aligned}

Answer: Option A

Explanation:

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

\end{aligned}