First slide

Permutations

Question

A seven digit number divisible by 9 is to be formed by using 7 out of numbers ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ . The number of ways in which this can be done is:

Moderate

A

7!
B

$2 (7)!$
C

$3 (7)!$
D

$4.7!$

Solution

Sum of 7 digits = a multiple of 9
We know sum of numbers 1,2,3,4,5,6,7,8,9 is 45

So, two left numbers should also have sum as 9. The pairs to be left are (1, 8) (2, 7) (3, 6), (4, 5) with each pair left number of 7 digit numbers is 7!

So with all 4 pairs = $4 \times 7!$

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