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The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only, is
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The correct answer is:
77
77
There are two possible cases arise:
Case I Five 1's, one 2's, one 3's
$\therefore$ Number of numbers $=\frac{7 !}{5 !}=42$
Case II Four 1's, three 2's
$\therefore$ Number of numbers $=\frac{7 !}{4 ! 3 !}=35$
$\therefore$ Total number of numbers $=42+35=77$
Case I Five 1's, one 2's, one 3's
$\therefore$ Number of numbers $=\frac{7 !}{5 !}=42$
Case II Four 1's, three 2's
$\therefore$ Number of numbers $=\frac{7 !}{4 ! 3 !}=35$
$\therefore$ Total number of numbers $=42+35=77$
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