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Question

Three six faced dice are. thrown together. The probability that the sum of the numbers appearing on the dice is k (3k8) is

A
k(k1)432
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B
k2432
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C
(k1)(k2)432
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D
None of these
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Solution

The correct option is C (k1)(k2)432
Total number of ways =63=216
Number of favorable ways = coefficients of xk in (x+x2+...+x6)3
= coefficients of xk3 in (1+x+x2+...+x6)3
= coefficients of xk3 in (1x6)3(1x)3
Since 3k80k35, we have numner of favorable ways
= coefficients of xk3 in (1x)3
=k3+31Ck3=k1C2
Required probability =k1C2216=(k1)(k2)432

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