
A seven digit number divisible by \[9\] is to be formed by using \[7\]out of number \[\left\{ {1,2,3,4,5,6,7,8,9} \right\}\]. The number of ways in which this can be done is ?
A \[7!\]
B \[2 \times 7!\]
C \[3 \times 7!\]
D \[4 \times 7!\]
Answer
161.4k+ views
Hint: A number is divisible by \[9\] when digit sum of the number is \[9\]. To find number of ways of \[7\]digit number divisible by \[9\], find sum of all digits in the set. As there \[9\]digits, find the pairs that can be removed from the set. Then check number of ways of \[7\]digit number divisible by \[9\],
Formula Used:\[n! = 1 \times 2 \times 3 \times ... \times n\]
Here n is any natural number.
\[{}^n{C_r} = \dfrac{{n!}}{{\left[ {r!\left( {n - r} \right)!} \right]}}\]
Here n is number of items in the collection and r is the number if item selected.
Complete step by step solution:The given data is seven-digit number divisible by \[9\] which is form using \[7\]digits from the set \[\left\{ {1,2,3,4,5,6,7,8,9} \right\}\].
First we need to understand the number divisible by \[9\]is multiplier of \[9\]. We can write it as follows.
Sum of \[7\]digits \[ = 9m\] Where m be any natural number.
Now find the sum of all numbers in the set \[\left\{ {1,2,3,4,5,6,7,8,9} \right\}\].
\[1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45\]
Now, \[45\]is multiplier of \[9\]. So we need to take out a pair from given set whose digit sum is \[9\] so that sum of remaining \[7\] digits will be multiple of \[9\].
Find the possible pairs to remove from set.
\[\left( {1,8} \right),\left( {2,7} \right),\left( {3,6} \right),\left( {4,5} \right)\]
For every pair there are \[7\]digit number and we have \[4\]such pair.
\[\begin{array}{l}{}^4{C_3} = \dfrac{{1 \times 2 \times 3 \times 4}}{{1 \times 2 \times 3}}\\{}^4{C_3} = 4\end{array}\]
Find the number of ways of \[7\]digit numbers which is divisible by \[9\].
Total number of ways \[ = 4 \times 7!\]
Option ‘D’ is correct
Note: Here, after eliminating each pair there are \[7\]digits and these can be arranged in \[7!\]ways not only \[7\]. Also, for pair section we cannot select\[{}^4{C_1}\]. These are the common mistake happens from the students.
Formula Used:\[n! = 1 \times 2 \times 3 \times ... \times n\]
Here n is any natural number.
\[{}^n{C_r} = \dfrac{{n!}}{{\left[ {r!\left( {n - r} \right)!} \right]}}\]
Here n is number of items in the collection and r is the number if item selected.
Complete step by step solution:The given data is seven-digit number divisible by \[9\] which is form using \[7\]digits from the set \[\left\{ {1,2,3,4,5,6,7,8,9} \right\}\].
First we need to understand the number divisible by \[9\]is multiplier of \[9\]. We can write it as follows.
Sum of \[7\]digits \[ = 9m\] Where m be any natural number.
Now find the sum of all numbers in the set \[\left\{ {1,2,3,4,5,6,7,8,9} \right\}\].
\[1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45\]
Now, \[45\]is multiplier of \[9\]. So we need to take out a pair from given set whose digit sum is \[9\] so that sum of remaining \[7\] digits will be multiple of \[9\].
Find the possible pairs to remove from set.
\[\left( {1,8} \right),\left( {2,7} \right),\left( {3,6} \right),\left( {4,5} \right)\]
For every pair there are \[7\]digit number and we have \[4\]such pair.
\[\begin{array}{l}{}^4{C_3} = \dfrac{{1 \times 2 \times 3 \times 4}}{{1 \times 2 \times 3}}\\{}^4{C_3} = 4\end{array}\]
Find the number of ways of \[7\]digit numbers which is divisible by \[9\].
Total number of ways \[ = 4 \times 7!\]
Option ‘D’ is correct
Note: Here, after eliminating each pair there are \[7\]digits and these can be arranged in \[7!\]ways not only \[7\]. Also, for pair section we cannot select\[{}^4{C_1}\]. These are the common mistake happens from the students.
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