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Determine how many solutions there are to the equation \mathrm{x+y+z=29}, where \mathrm{\mathrm{x}, \mathrm{y}}, and \mathrm{\mathrm{z}} are all \mathrm{\mathrm{W}}.

Option: 1

378


Option: 2

556


Option: 3

676


Option: 4

675


Answers (1)

best_answer

Given that \mathrm{x+y+z=29}
For \mathrm{x_{1}+x_{2}+x_{3}+\ldots \ldots x_{r}=n}

The number of solutions will be \mathrm{={ }^{(n-1)} C_{(r-1)}}
Then for \mathrm{x+y+z=29}
 

The number of solutions is given as \mathrm{={ }^{29-1} C_{3-1}}
                                                         \mathrm{={ }^{28} C_{2}}
                                                         \mathrm{=378}
The correct option is a

Posted by

Gautam harsolia

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