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 C

Explanation:

We need to find the ways that vowels NEVER come together.
Vowels are A, E
Let the word be FTR(AE) having 4 words.
Total ways = 4! = 24
Vowels can have total ways 2! = 2
Number of ways having vowel together = 48
Total number of words using all letter = 5! = 120

Number of words having vowels never together = 120-48
= 72

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 =

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

Answer: Option D

Explanation:

In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.
So we can have

(four boys) or (three boys and one girl) or (two boys and two girls) or (one boy and three gils)

This combination question can be solved as

$$\begin{aligned} (^{6}{C}_{4}) + (^{6}{C}_{3} * ^{4}{C}_{1}) + \\ + (^{6}{C}_{2} * ^{4}{C}_{2}) + (^{6}{C}_{1} * ^{4}{C}_{3}) \\ = \left[\dfrac{6 \times 5 }{2 \times 1}\right] + \left[\left(\dfrac{6 \times 5 \times 4 }{3 \times 2 \times 1}\right) \times 4\right] + \\\left[\left(\dfrac{6 \times 5 }{2 \times 1}\right)\left(\dfrac{4 \times 3 }{2 \times 1}\right)\right] + \left[6 \times 4 \right] \\ = 15 + 80 + 90 + 24\\ = 209 \end{aligned}$$

Answer: Option B

Explanation:

As we can see the letter "E" is twice in given word, so Required Number

$$\begin{aligned} = \frac{7!}{2!} \\ = \frac{7*6*5*4*3*2!}{2!} \\ = 2520 \end{aligned}$$