The basic case is just (m+n) choose m, since you're picking positions for the 'a's and the 'b's fill the rest.
Start by deriving the formula for two characters using the multinomial coefficient: total permutations of m+n items divided by the indistinguishable permutations of each character. Then generalize to k characters by extending the same logic: N! divided by the product of factorials of each count. Justify each step by explaining the division principle and the multiplication principle.
Pro tip: Emphasize that this is the multinomial coefficient and connect it to the binomial coefficient when k=2. Also mention that the formula assumes all characters of the same type are identical, which is key to the division by factorials.
Clarify that we are counting distinct arrangements of m identical 'a's and n identical 'b's, and later k distinct characters with given counts.
Treat all m+n positions as distinct initially, giving (m+n)! permutations. Then divide by m! and n! to account for identical 'a's and 'b's, yielding (m+n)!/(m! n!).
Extend the same reasoning: total permutations of N distinct items is N!, then divide by the product of factorials of each count to remove overcounting from identical items. The formula is N!/(c1! c2! ... ck!).
Explain that each distinct string is counted exactly ∏(ci!) times in the N! permutations, so dividing by that product gives the correct count. This is the multinomial coefficient.
Illustrate with small numbers (e.g., m=2, n=1 gives 3 strings) and mention that if any ci=0, the term is 1, and the formula still holds.
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