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An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is:
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Verified Answer
The correct answer is:
$36(7 !)$
$36(7 !)$
We know that any number is divisible by 9 if sum of the digits of the number is divisible by 9 .
Now sum of the digits from 0 to 9
$$
\begin{aligned}
&=0+1+2+3+4+5+6+7+8+9 \\
&=45
\end{aligned}
$$
Hence to form 8 digits numbers which are divisible by 9 , a pair of digits either 0 and 9,1 and 8,2 and 7,3 and 6 or 4 and 5 are not used.
Hence total number of 8 digits numbers which are divisible by 9
$$
\begin{aligned}
&=8 \times(7 !)+7 \times(7 !)+7 \times(7 !)+7 \times(7 !) \\
&+7 \times(7 !) \\
&=36 \times(7 !)
\end{aligned}
$$
Now sum of the digits from 0 to 9
$$
\begin{aligned}
&=0+1+2+3+4+5+6+7+8+9 \\
&=45
\end{aligned}
$$
Hence to form 8 digits numbers which are divisible by 9 , a pair of digits either 0 and 9,1 and 8,2 and 7,3 and 6 or 4 and 5 are not used.
Hence total number of 8 digits numbers which are divisible by 9
$$
\begin{aligned}
&=8 \times(7 !)+7 \times(7 !)+7 \times(7 !)+7 \times(7 !) \\
&+7 \times(7 !) \\
&=36 \times(7 !)
\end{aligned}
$$
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