In how many ways can the letters of the CHEATER be arranged

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 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 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 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 B

Explanation:

This question seems to be a bit typical, isn't, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways

Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24

Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.