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 :
Solution
(I) First possibility is 1, 1, 1, 1, 1, 2, 3<br><br>required number = ${{7!} \over {5!}}$ = 7 $\times$ 6 = 42<br><br>(II) Second possibility is 1, 1, 1, 1, 2, 2, 2<br><br>required number = ${{7!} \over {4!3!}} = {{7 \times 6 \times 5} \over 6} = 35$<br><br>Total = 42 + 35 = 77
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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