A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours

Answer: Option D

Explanation:

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

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:

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 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}