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Let E be the event that seven digit integers formed using the digits 1, 2, 3 and 4 only and sum of digits equal to 12. Also, let S be the event that all the seven digit numbers are formed using 1, 2, 3 and 4 only. The $P (E) . n (S) =_____.$

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answer is 413.

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Detailed Solution

(413)
$P (E) = \frac{n (E)}{n (s)}$
$∴$ P(E).n(S) = n(E) = no. of seven digit nos formed by 1,2,3 and 4 only such that sum is 12
$= 5 + 7 - 1_{C_{7 - 1}} - \frac{7!}{6!} - \frac{7!}{5!}$ = 413

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