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  •  There are a total of _______ 3-digit numbers with digit sums of 20.Option: 1 210Option:

 There are a total of _______ 3-digit numbers with digit sums of 20.

Option: 1

210


Option: 2

280


Option: 3

209


Option: 4

115


Answers (1)

best_answer

Assuming the 3-digit number as xyz.

Given that sum of the digits of the number is 10
Then \mathrm{x+y+z=20}

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=20}
\mathrm{s+y+z=19}

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

Then the number of solutions will be
\mathrm{={ }^{19+3-1} C_{3-1}}
\mathrm{=^{21} C_{2}}
\mathrm{=210}

But \mathrm{x=20} is not possible for \mathrm{s=19}
The total solutions will be \mathrm{=210-1}
                                         \mathrm{=209}

The correct option is (c)

Posted by

vinayak

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