Question Detail
In how many words can be formed by using all letters of the word BHOPAL
- 420
- 520
- 620
- 720
Answer: Option D
Explanation:
Required number
\begin{aligned}
= 6! \\
= 6*5*4*3*2*1 \\
= 720
\end{aligned}
1. Evaluate combination
\begin{aligned}
^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\
\end{aligned}
- 161700
- 151700
- 141700
- 131700
Answer: Option A
Explanation:
\begin{aligned}
^{n}{C}_r = \frac{n!}{(r)!(n-r)!} \\
^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\
= \frac{100*99*98*97!}{(97)!(3)!} \\
= \frac{100*99*98}{3*2*1} \\
= \frac{100*99*98}{3*2*1} \\
= 161700
\end{aligned}
2. Evaluate combination
\begin{aligned}
^{100}{C}_{100}
\end{aligned}
- 10000
- 1000
- 10
- 1
Answer: Option D
Explanation:
\begin{aligned}
^{n}{C}_{n} = 1 \\
^{100}{C}_{100} = 1
\end{aligned}
3. 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
- 64
- 128
- 132
- 222
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}
4. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours
- 12
- 24
- 48
- 168
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.
5. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done
- 456
- 556
- 656
- 756
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}