posted by Ryan .
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 probaility that this number is divisible by 11.
haha is this a last minute calendar problem by any chance? because it's october tomorrow and i have have this problem too! lets hope somebody answers soon...
The digit combinations of a 5-digit number having a sum of 43 is quite limited.
The digits are 99997, 99988.
99997 can place the 7 in 5 positions, so the number of outcomes using 4-9's and a 7 is 5.
Using 3-9's and 2-8's has 5!/(3!2!)=10 ways of arranging the numbers.
Of these, only 98989, 97999 and 99979 are divisible by 11.
Can you continue?
Check my thinking.
Yes seth it is