A bag contains four tickets numbered 00, 01, 10, 11. Four tickets are chosen at random with replacement, the probability that sum of the numbers on the tickets is 23, is
A
332
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B
164
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C
5256
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D
7256
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Solution
The correct option is A332 The total number of ways of choosing the tickets is 4×4×4×4=256
The number of ways in which the sum can be 23.
= coefficients of x23 in (1+x+x10+x11)4
= coefficients of x23 in (1+x)4(1+x10)4
= coefficients of x23 in (1+4x+6x2+4x3+x4)×(1+4x10+6x20) =4×6=24
The probability of required event =24256=332