Trials and Tribulations in Permutations and Combinations

This problem has a systematic approach as follows :

let the 5 digits be a, b ,c ,d & e.

Given, a + b + c + d + e = 41 ----->1.

For multiple of 11 we should have a + c + e - b - d = 0 or multiple of 11 ------>2.

Substituting 1 in 2, we have 41 - 2b - 2d = 11 -------->3.

We cannot have 0 in above equation because 41 is an odd number, but we have (2b + 2d), which is not possible.We cannot have 22 in the same way and 33 because (b + d) becomes 4,but (a + c + e) becomes 37 which cannot form single digit numbers.

The equation 3 becomes b + d = 15. & a + c + e = 26.

(b,d) can be (9,6) , (6,9) , (8,7) & (7,8) which is 4 ways.

(a,c,e) can be (9,9,8) , (9,8,9) , (8,9,9) which is 3 ways.

So, total combinations is 4 * 3 = $\boxed{12}$ ways.

There are, in total, 5 conditions which satisfy the problem, id est :- 99995,99986,99977,99887,98888. Out of all these only 99986 and 99887 satisfy the second condition. Now, we apply the divisibility rule of 11. Therefore, No. of numbers that fall under the category of 99986 satisfying the second condition are 3!/2! * 2! and no. of numbers that fall under the category of 99887 satisfying the second condition are 3!/2! * 2! . Add them.