In how many way the letter of the word "APPLE" can be arranged

Answer: Option A

Explanation:

From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.

Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]

\begin{aligned}
= 1 + \left[3 \times 6 \right] + \left[3 \times \left(\dfrac{6 \times 5}{2 \times 1}\right) \right]
= 1 + 18 + 45
= 64
\end{aligned}

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 B

Explanation:

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

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}