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Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, y/9x =
- a)25
- b)15
- c)5
- d)10
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Words of length 10 are formed using the letters A, B, C, D, E, F, G, H...
Ans. 5
x = 10!
Hence option. c
Most Upvoted Answer
Words of length 10 are formed using the letters A, B, C, D, E, F, G, H...
Solution:
Number of ways to form words of length 10 using the given letters = 10!
$x$: Number of words where no letter is repeated
In this case, the first letter can be any of the 10 letters. The second letter can be any of the remaining 9 letters. Similarly, the third letter can be any of the remaining 8 letters, and so on. Hence, the number of such words is:
$x = 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 = 10!$
$y$: Number of words where exactly one letter is repeated twice and no other letter is repeated
In this case, we can choose the letter to be repeated in $\binom{10}{1}$ ways. Once we have chosen the letter, we can choose the positions for the two occurrences of the chosen letter in $\binom{10}{2}$ ways. The remaining 8 letters can be arranged in the remaining 8 positions in $8!$ ways. Hence, the number of such words is:
$y = \binom{10}{1}\times \binom{10}{2}\times 8!$
Therefore, $\frac{y}{9x}=\frac{\binom{10}{1}\times \binom{10}{2}\times 8!}{9\times 10!}=\frac{5}{126}$
$\frac{y}{9x}=\frac{5}{126}=\frac{a}{100}$
Hence, $a=5$.
Therefore, the correct option is (C) 5.
Number of ways to form words of length 10 using the given letters = 10!
$x$: Number of words where no letter is repeated
In this case, the first letter can be any of the 10 letters. The second letter can be any of the remaining 9 letters. Similarly, the third letter can be any of the remaining 8 letters, and so on. Hence, the number of such words is:
$x = 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 = 10!$
$y$: Number of words where exactly one letter is repeated twice and no other letter is repeated
In this case, we can choose the letter to be repeated in $\binom{10}{1}$ ways. Once we have chosen the letter, we can choose the positions for the two occurrences of the chosen letter in $\binom{10}{2}$ ways. The remaining 8 letters can be arranged in the remaining 8 positions in $8!$ ways. Hence, the number of such words is:
$y = \binom{10}{1}\times \binom{10}{2}\times 8!$
Therefore, $\frac{y}{9x}=\frac{\binom{10}{1}\times \binom{10}{2}\times 8!}{9\times 10!}=\frac{5}{126}$
$\frac{y}{9x}=\frac{5}{126}=\frac{a}{100}$
Hence, $a=5$.
Therefore, the correct option is (C) 5.
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Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then,y/9x=a)25b)15c)5d)10Correct answer is option 'C'. Can you explain this answer?
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