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An $11$ letter word has:

$$4 \text{ A's}\\3 \text{ N's}\\2\text{ G's}\\ 1\text{ M}\\1\text{ T}$$

Find the number of distinct combinations of the word such that there are no A’s in the first six places?

I initially thought to find the number of choices that have A's in the first six places and then to subtract this from the total number of distinct combinations, but there is a flaw in my calculations somewhere.

$$^6C_4 = 15$$which is the number of choices for A's in the first six places.

$$\frac{7!}{3!\times 2}=420$$

$$\text{Total}=\frac{11!}{4!\times3!\times2}-420\times15=132300$$

This just seems too unreasonable, would someone be able to tell me where I have gone wrong in my calculations?

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No A's in the first six places means that they must all go in the last five places; there are $\binom54=5$ ways of arranging the A's as such. The remaining $7$ letters can be arranged in $\frac{7!}{3!2!1!1!}=420$ ways, so there are $5×420=2100$ valid arrangements.

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You've calculated the number of ways all the As are in the first 6 places and subtracted them from the total number, rather than calculating the number of ways at least one A is in the first 6 which is what you were aiming for.

A better approach would be to do a similar thing, but count all the ways the As are in the last 5 places, i.e $\binom{5}{4}$, and multiply by the number of ways of ordering the rest of the characters, $\frac{7!}{3!2!}$ to get $2100$ ways in total.

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