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Question Detail
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
1. 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}
2. Evaluate permutation equation
\begin{aligned} ^{75}{P}_2\end{aligned}
- 5200
- 5300
- 5450
- 5550
Answer: Option D
Explanation:
\begin{aligned}
^n{P}_r = \frac{n!}{(n-r)!} \\
^{75}{P}_2 = \frac{75!}{(75-2)!} \\
= \frac{75*74*73!}{(73)!} \\
= 5550
\end{aligned}
3. How many words can be formed from the letters of the word "AFTER", so that the vowels never comes together.
- 48
- 52
- 72
- 120
Answer: Option C
Explanation:
We need to find the ways that vowels NEVER come together.
Vowels are A, E
Let the word be FTR(AE) having 4 words.
Total ways = 4! = 24
Vowels can have total ways 2! = 2
Number of ways having vowel together = 48
Total number of words using all letter = 5! = 120
Number of words having vowels never together = 120-48
= 72
4. How many words can be formed by using all letters of TIHAR
- 100
- 120
- 140
- 160
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}
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}
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