From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done

Answer: Option D

Explanation:

Lets suppose there were x number os team in cricket cup then from the given statement we can calculate :
\begin{aligned}
=> ^n C _2 = 153 \\
\text{ it given n = 18 and n = -17} \\
\end{aligned}
As answer cannot be negative to 18 is the answer

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 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}

Answer: Option D

Explanation:

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
So we can have
(5 men) or (4 men and 1 woman) or (3 men and 2 woman)

\begin{aligned}
(^{5}{C}_{5}) + (^{5}{C}_{4} * ^{6}{C}_{1}) + \\
+ (^{5}{C}_{3} * ^{6}{C}_{2}) \\

= \left[\dfrac{7 \times 6 }{2 \times 1}\right] + \left[\left( \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right) \times 6 \right] + \\ \left[\left( \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right) \times \left( \dfrac{6 \times 5}{2 \times 1} \right) \right] \\
= 21 + 210 + 525 = 756
\end{aligned}

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