In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there

Answer: Option D

Explanation:

\begin{aligned}
^n{P}_r = \frac{n!}{(n-r)!} \\
^{75}{P}_2 = \frac{75!}{(75-2)!} \\
= \frac{75*74*73!}{(73)!} \\
= 5550


\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 A

Explanation:

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

\end{aligned}

Answer: Option A

Explanation:

word SIGNATURE contains total 9 letters.
There are four vowels in this word, I, A, U and E

Make it as, SGNTR(IAUE), consider all vowels as 1 letter for now
So total letter are 6.
6 letters can be arranged in 6! ways = 720 ways
Vowels can be arranged in themselves in 4! ways = 24 ways

Required number of ways = 720*24 = 17280

Answer: Option B

Explanation:

Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)

This has 7 lettes, where R is twice so value = 7!/2!
= 2520

Vowel O is 3 times, so vowels can be arranged = 5!/3!

= 20

Total number of words = 2520 * 20 = 50400