How many 4-digit permutations can be formed out of:1, 1, 1, 2, 2, 3, 4, 5. Thanks for help.
It is much easier to find the "combinations" first and then calculate the permutations from those combinations:
There are: {1, 1, 1, 2}, {1, 1, 1, 3}, {1, 1, 1, 4}, {1, 1, 1, 5}, {1, 1, 2, 2}, {1, 1, 2, 3}, {1, 1, 2, 4}, {1, 1, 2, 5}, {1, 1, 3, 4}, {1, 1, 3, 5}, {1, 1, 4, 5}, {1, 2, 2, 3}, {1, 2, 2, 4}, {1, 2, 2, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 2, 3, 4}, {2, 2, 3, 5}, {2, 2, 4, 5}, {2, 3, 4, 5} = 22 combinations.
Four of them have =4!/3! x 4 =16 permutations
One of them has =4!/2!2! =6 p
Twelve of them have =4!/2! x 12 =144 p
Five of them have =4! x 5 =120 p
So: 16 + 6 + 144 + 120 =286 permutations.
