The number of $6 - d i g i t$ numbers that can be formed using the digits $1, 3, 5$ so that $5$ occurs twice in each number, is $x y z$ then the value of $(x + y + z)$ is equal to

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Solution

Total No. $6$ digits
--------------------
$^{6} C_{2} \times 1$
Refer image.
$∴^{6} C_{2} \times 1 \times 2 \times 2 \times 2 \times 2$

$\Rightarrow \frac{6!}{2! 4!} \times 2^{4}$

$\Rightarrow \frac{6 \times 54!}{2! 4!} \times 2^{4}$

$\Rightarrow 30 \times 8 = 240 = x y z$
$x = 2$
$y = 4$
$z = 0$ .
$x + y + z = 2 + 4 + 0 = 6$

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The number of 6−digit numbers that can be formed using the digits 1,3,5 so that 5 occurs twice in each number, is xyz then the value of (x+y+z) is equal to

Total No. 6 digits--------------------6C2×1Refer image.∴6C2×1×2×2×2×2⇒6!2!4!×24⇒6×54!2!4!×24⇒30×8=240=xyzx=2y=4z=0.x+y+z=2+4+0=6

The number of $6 - d i g i t$ numbers that can be formed using the digits $1, 3, 5$ so that $5$ occurs twice in each number, is $x y z$ then the value of $(x + y + z)$ is equal to