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There are _______ three-digit numbers with digit sums of 32.

Option: 1

678


Option: 2

561


Option: 3

451


Option: 4

560


Answers (1)

best_answer

Assuming the 3-digit number as xyz.
Given that sum of the digits of the number is 32

Then \mathrm{x+y+z=32}

Two digits can be zero, but one of them has to be greater than 1.
Assuming that \mathrm{x \geq 1, y \geq 0} and \mathrm{z \geq 0}

Again assuming that \mathrm{x-1=s}
\mathrm{x=1+s} Then
\mathrm{1+s+y+z=32}
\mathrm{s+y+z=31}

And
\mathrm{s \geq 0, y \geq 0} and \mathrm{z \geq 0}

Then the number of solutions will be
\mathrm{=^{32+3-1} C_{3-1}}
\mathrm{={ }^{34} C_{2}}
\mathrm{=561}

But \mathrm{x=32} is not possible for \mathrm{s=31}

The total solutions will be \mathrm{=561-1}
                                         \mathrm{=560}

The correct option is (d)

Posted by

Nehul

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