JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 5)
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 :
35
42
82
77
Explanation
(I) First possibility is 1, 1, 1, 1, 1, 2, 3
required number = $${{7!} \over {5!}}$$ = 7 $$\times$$ 6 = 42
(II) Second possibility is 1, 1, 1, 1, 2, 2, 2
required number = $${{7!} \over {4!3!}} = {{7 \times 6 \times 5} \over 6} = 35$$
Total = 42 + 35 = 77
required number = $${{7!} \over {5!}}$$ = 7 $$\times$$ 6 = 42
(II) Second possibility is 1, 1, 1, 1, 2, 2, 2
required number = $${{7!} \over {4!3!}} = {{7 \times 6 \times 5} \over 6} = 35$$
Total = 42 + 35 = 77
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