SOLUTION: If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to 43, compute the probability that this number will be divisible b

Three of the five digits must be 9's because if we had fewer 9's, say even as
many as two 9's, and the rest as large as they could possibly be, (that is, if
they were all three 8's) you'd only have sum of digits 9+9+8+8+8=42 which is 1
short of 43. So the three 9's each integer must have makes up 27 of the
required total of 43, leaving 16 for the remaining two digits to have as a
sum.  The only way for two digits to have sum 16 is 9+7 and 8+8. Therefore all
possible such integers must be arrangements of the digits of either 79999 or of
88999.

I.  Count the arrangements of the digits of the integer 79999.

There are 5 places to put the 7.  So there are 5 arrangements of the digits of
79999.  Since there are so few we can list these.

1.  79999
2.  97999 = 11*8909
3.  99799 
4.  99979 = 11*9089
5.  99997

Only two of these five are divisible by 11, 

II.  Count the arrangements of the digits of the integer 88999.

There are "5 choose 2" or 5C2 or  places to put the two
8's.  So there are 10 arrangements of the 
digits of 88999.  That's not too many to list either.  Listing them:

 1. 88999
 2. 89899
 3. 89989
 4. 89998
 5. 98899
 6. 98989 = 11*8999
 7. 98998 
 8. 99889
 9. 99898
10. 99988

We find that only one of these is divisible by 11.

So of the 5+10 or 15 positive integers with sum of digits 43,
exactly 3 of them are divisible by 11.

So the desired probability is "3 out of 15" or  or  

Edwin