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The number of integers $x, y, z, w$ satisfying $x+y+z+w=25$ and $x, y, z \geq-1, w \geq 1$, is
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The correct answer is:
${ }^{30} C_3$
It is given that, $x, y, z \geq-1$ and $w \geq 1$
Now, let $a=x+1 \geq 0$
$b=y+1 \geq 0$
$c=z+1 \geq 0$
$d=w-1 \geq 0$
$\therefore$ The given equation $x+y+z+w=25$ reduce to
$(a-1)+(b-1)+(c-1)+(d+1)=25$
$\Rightarrow \quad a+b+c+d=27$
$\therefore$ The number of required solution is same as the number of non-negative integral solutions of equation
$a+b+c+d=27$
and it is equal to ${ }^{27+4-1} C_{4-1}={ }^{30} C_3$
Now, let $a=x+1 \geq 0$
$b=y+1 \geq 0$
$c=z+1 \geq 0$
$d=w-1 \geq 0$
$\therefore$ The given equation $x+y+z+w=25$ reduce to
$(a-1)+(b-1)+(c-1)+(d+1)=25$
$\Rightarrow \quad a+b+c+d=27$
$\therefore$ The number of required solution is same as the number of non-negative integral solutions of equation
$a+b+c+d=27$
and it is equal to ${ }^{27+4-1} C_{4-1}={ }^{30} C_3$
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