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Question Detail
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
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. Evaluate \begin{aligned}
\frac{30!}{28!}
\end{aligned}
- 970
- 870
- 770
- 670
Answer: Option B
Explanation:
\begin{aligned}
= \frac{30!}{28!} \\
= \frac{30 * 29 * 28!}{28!} \\
= 30 * 29 = 870
\end{aligned}
3. 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}
4. 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}
5. Evaluate permutation equation
\begin{aligned} ^{59}{P}_3 \end{aligned}
- 195052
- 195053
- 195054
- 185054
Answer: Option C
Explanation:
\begin{aligned}
^n{P}_r = \frac{n!}{(n-r)!} \\
^{59}{P}_3 = \frac{59!}{(56)!} \\
= \frac{59 * 58 * 57 * 56!}{(56)!} \\
= 195054
\end{aligned}
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