The number of integral solutions to the equat | vedclass

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(B) Given the equation $x+y+z=21$ with constraints $x \geq 1, y \geq 3, z \geq 4$.Let $x' = x-1, y' = y-3, z' = z-4$,where $x', y', z' \geq 0$.Substit

Vedclass · Accepted Answer

$105$

Vedclass · Answer

(B) Given the equation $x+y+z=21$ with constraints $x \geq 1, y \geq 3, z \geq 4$.Let $x' = x-1, y' = y-3, z' = z-4$,where $x', y', z' \geq 0$.Substituting these into the equation: $(x'+1) + (y'+3) + (z'+4) = 21$.$x' + y' + z' + 8 = 21$.$x' + y' + z' = 13$.The number of non-negative integral solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n=13$ and $r=3$.Number of solutions = $\binom{13+3-1}{3-1} = \binom{15}{2}$.$\binom{15}{2} = \frac{15 	imes 14}{2 	imes 1} = 15 	imes 7 = 105$.

The number of integral solutions to the equation $x+y+z=21$,where $x \geq 1, y \geq 3, z \geq 4$,is equal to $..........$.

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