Question:
The number of 7 digits integers formed by using the digits 2, 3, 4, 5 only and having the sum of digits equal to 18 is ________.
The number of 7 digits integers formed by using the digits 2, 3, 4, 5 only and having the sum of digits equal to 18 is ________.
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Subtract 2 from every digit so each digit becomes 0 to 3 with total 4, then count solutions and remove those with a digit exceeding 3.
Updated On: Nov 12, 2026
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Correct Answer: 203
Solution and Explanation
Let the digits be \(d_1,\dots,d_7\) with each \(d_i\in\{2,3,4,5\}\) and \(\sum d_i=18\).
Put \(y_i=d_i-2\). Then \(y_i\in\{0,1,2,3\}\) and \(\sum y_i=18-14=4\).
Non-negative solutions of \(y_1+\dots+y_7=4\): \(\binom{4+6}{6}=\binom{10}{6}=210\).
Remove solutions in which some \(y_i\ge 4\): this is only possible if one \(y_i=4\) and the rest are 0, giving 7 solutions.
Required number \(=210-7=203\).
Answer: 203
Put \(y_i=d_i-2\). Then \(y_i\in\{0,1,2,3\}\) and \(\sum y_i=18-14=4\).
Non-negative solutions of \(y_1+\dots+y_7=4\): \(\binom{4+6}{6}=\binom{10}{6}=210\).
Remove solutions in which some \(y_i\ge 4\): this is only possible if one \(y_i=4\) and the rest are 0, giving 7 solutions.
Required number \(=210-7=203\).
Answer: 203
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