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. Evaluate permutation
\begin{aligned}
^5{P}_5
\end{aligned}

Answer: Option A

Explanation:

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

\end{aligned}

2. 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}

3. 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}

4. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw

Answer: Option A

Explanation:

From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.

Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]

\begin{aligned}
= 1 + \left[3 \times 6 \right] + \left[3 \times \left(\dfrac{6 \times 5}{2 \times 1}\right) \right]
= 1 + 18 + 45
= 64
\end{aligned}

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

2520
480
360
180

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}

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

1. Evaluate permutation \begin{aligned} ^5{P}_5 \end{aligned}

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

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

4. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw

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

1. Evaluate permutation
\begin{aligned}
^5{P}_5
\end{aligned}

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