How many words can be formed from the letters of the word "SIGNATURE" so that vowels always come together.

Answer: Option B

Explanation:

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

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 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:

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}