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Question Detail
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}
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. 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}
3. 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}
4. 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
5. 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}
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