Solution below.

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This is known as a stars and bars problem. I think of it like sticks and stones. We’re also into the territory of compositions and partitions.

Imagine 2 stones and 2 sticks. I want to insert the sticks between the stones, making small groups of stones. All the stones are identical, all the sticks are identical. I do care about the ordering though. 2 stones followed by 2 sticks is not the same as 2 sticks followed by 2 stones. One possibility is olol. Stone, stick, stone, stick. What’s in front of the first stick? 1 stone. What’s between the 2 sticks? 1 stones. What’s after the last stick? 0 stones. So this is a way to write 110. This again is a way to write a*b. 1 a, 1 b, 0 c’s. And a*b corresponds to our ab term above. (2ab.) How many terms are there? Well, in how many ways can we place the sticks? This is the same as saying, we have 2+2 positions and 2 sticks. In how many ways can we place the sticks? 4*3/2*1 = 6.

Each stone is a bit of a term. Each term has 2 important bits. Each bit can be a, b or c. 3 groups. To separate our a’s, b’s and c’s, we need 2 sticks. 1 stick between the a’s and the b’s, 1 between the b’s and the c’s. So, 3 letters becomes 3-1 sticks. And the exponent, 2, becomes 2 stones.

6 letters and the exponent 5 becomes 5 sticks and 5 stones becomes 10 positions and 5 sticks. In how many ways can we place the sticks? 10*9*8*7*6/5*4*3*2*1 = 252, the solution.