Given the digits (1,1,2,2,2,5):
a) How many 6 digit numbers can be formed?
We can place the two 1's in any 2 of the 6 positions.
That's C(6,2) = 15 ways.
For each of those 15 ways,
we can place the three 2's in any 3 of the remaining 4 positions.
That's C(4,3) = 4 ways.
We then place the 5 in the only 1 remaining position. (1 way)
Answer: 15×4×1 = 60 ways.
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b) How many 6 digit numbers are divisible by 5?
They must all end in 5.
We place the 5 as the 6th digit. (1 way)
We can place the two 1's in any 2 of the remaining 5 positions.
That's C(5,2) = 10 ways.
We then place the three 2's in the only 3 remaining positions. (1 way).
Answer: 1×10×1 = 10 ways.
Edwin