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Question Detail
In how many way the letter of the word "APPLE" can be arranged
- 20
- 40
- 60
- 80
Answer: Option C
Explanation:
Friends the main point to note in this question is letter "P" is written twice in the word.
Easy way to solve this type of permutation question is as,
So word APPLE contains 1A, 2P, 1L and 1E
Required number =
\begin{aligned}
= \frac{5!}{1!*2!*1!*1!} \\
= \frac{5*4*3*2!}{2!} \\
= 60
\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. 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}
3. 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}
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 ways can the letters of the word "CORPORATION" be arranged so that vowels always come together.
- 5760
- 50400
- 2880
- None of above
Answer: Option B
Explanation:
Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)
This has 7 lettes, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400
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