Question Detail

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:

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

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

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

20160
2520
360
80

Answer: Option B

Explanation:

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

2. Evaluate $\begin{aligned} \frac{30!}{28!} \end{aligned}$

Answer: Option B

Explanation:

$\begin{aligned} = \frac{30!}{28!} \\ = \frac{30 * 29 * 28!}{28!} \\ = 30 * 29 = 870 \end{aligned}$

3. Evaluate permutation equation
$\begin{aligned} ^{59}{P}_3 \end{aligned}$

195052
195053
195054
185054

Answer: Option C

Explanation:

$\begin{aligned} ^n{P}_r = \frac{n!}{(n-r)!} \\ ^{59}{P}_3 = \frac{59!}{(56)!} \\ = \frac{59 * 58 * 57 * 56!}{(56)!} \\ = 195054 \end{aligned}$

4. In how many way the letter of the word "APPLE" can be arranged

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 =
$\begin{aligned} = \frac{5!}{1!*2!*1!*1!} \\ = \frac{5*4*3*2!}{2!} \\ = 60 \end{aligned}$

5. How many words can be formed by using all letters of TIHAR

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

MATH@123 10 years ago

Sir, for the last Q cant we do it in this manner:
6 C 1 x 9 C 3 =504
(selecting one boy out of six and then having selected one, selecting 3 out the remaining nine)