In how many words can be formed by using all letters of the word BHOPAL

Answer: Option C

Explanation:

\begin{aligned}
^n{P}_r = \frac{n!}{(n-r)!} \\
^{59}{P}_3 = \frac{59!}{(56)!} \\
= \frac{59 * 58 * 57 * 56!}{(56)!} \\
= 195054
\end{aligned}

Answer: Option D

Explanation:

Required number
\begin{aligned}
= 6! \\
= 6*5*4*3*2*1 \\
= 720
\end{aligned}

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 A

Explanation:

\begin{aligned}
^{n}{C}_r = \frac{n!}{(r)!(n-r)!} \\
^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\
= \frac{100*99*98*97!}{(97)!(3)!} \\
= \frac{100*99*98}{3*2*1} \\
= \frac{100*99*98}{3*2*1} \\
= 161700
\end{aligned}