I have hard time counting the digits
Find the number of positive integers that satisfy the following conditions:
* Each digit is a 1 or a 2 or a 3
* The sum of the digits is 10
+118608 I would list out the possible number conbiations and then work out how many numbers I can make with each combination.
The maximum number of 3s is 3 that is 3331 there are 4 ways to order this
Now how many wars can there be two 3s
3,3,2,2 4!/(2!2!) = 6ways to order
3,3,1,1,2 5!/(2!2!)=30 ways to order
3,3,1,1,1,1, 6!/(2!4!) = 15 ways to order
How many ways can there be one 3s
3,2,2,2,1 5!/(3!) =
3,1,1,2,2,1 6!/(3!2!)=
3,1,1,1,1,2,1 7!/(5!)=
3,1,1,1,1,1,1,1 8!/7! =
Ok no more combinations use a 3
now go through a similar process with the 2s
starting with
2,2,2,2,2
etc
When you have finished add them all together.
