Question Detail
Evaluate permutation
\begin{aligned}
^5{P}_5
\end{aligned}
- 120
- 110
- 98
- 24
Answer: Option A
Explanation:
\begin{aligned}
^n{P}_n = n! \\
^5{P}_5 = 5*4*3*2*1 \\
= 120
\end{aligned}
1. In a Cricket cup total 153 matches were played and every two teams played exactly one match with each other. So what were the total number of teams participating in Cricket Cup ?
- 15
- 16
- 17
- 18
- 19
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
2. 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.
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. 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
- 109
- 128
- 138
- 209
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}
5. 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}