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A five digit number if formed by the digits 1,2,3,4,5,6 and 8. The probability that the number has even digit at both ends isA. `(3)/(7)`B. `(4)/(7)`C. `(2)/(7)`D. none of these |
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Answer» Correct Answer - C Total number of 5 digit numbers formed with the given digits `= .^(7)C_(5)xx 5!` There are 4 even digits out of which two even digits (for both ends) can be chosen in `.^(4)C_(2)` ways. Therefore, the two ends can be filled in `2xx .^(4)C_(2)` ways. Now, remaining three places can be filled in `.^(5)C_(3)xx3!` ways. `therefore` Number of 5 digit, numbers having even digits at both ends `=2xx .^(4)C_(2) xx .^(5)C_(3)xx3!` Required probability `=(2xx .^(4)C_(2)xx .^(5)C_(3)xx3!)/(.^(7)C_(5)xx5!)=(2)/(7)` |
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