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

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:

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

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 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:

As we can see the letter "E" is twice in given word, so Required Number
$$\begin{aligned} = \frac{7!}{2!} \\ = \frac{7*6*5*4*3*2!}{2!} \\ = 2520 \end{aligned}$$