Consider all arrangements of INSTRUCTOR. How many have N left of C and also left of S?

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Is there any way to solve this without the casework?

The casework way I have in mind is with $9 \choose 2,2$ for when you put the N in the first spot and then the 9 remaining letters in the remaining spots, and then $7*{8\choose2,2}$ for when you put the N in the second spot and pick one letter from the 7 (excluding C and S) in front.Then, I'm unsure how to proceed as the two Rs and two Ts make it difficult to count the number of ways to put N in the third spot.

I'd appreciate any help. Thanks!

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What you have might work, but I think the following is more slick. Choose places for the N,C and S before filling in anything else. Put the N in the left most of the three in one way. In the remaining two spots we can put the C and S in in two ways. Now fill in the rest of the spots using the remaining letters.