thanks for y'all's help. i'm stumped on another. how many distinct permutations can be formed using the letters in TALLAHASSEE?
Using all the letters?
yes
Ok so the first thing to realize is that you cannot distinguish between: TALLAHASSEE and TALLAHASSEE (where I just flipped the last two E's)
So rather than looking at his as a permutation, think instead about having a collection of distinct letters: {T, A, L, H, S, E} Where you need to choose where to place one T, then 3 As, then 2 Ls, then one H, then 2 Ss, then 2 Es
3 A's, 2 L's , 2 S's , 2 E's , 1 H, 1 T total letters = 11 hhmm - i'll have to check my stats..
There are 11 spots in which you can place the letters. So lets start with the first letter in our distinct collection: T We need to place one T in the set of 11 spots \(11 \choose 1\) Then we need to place 3 As: \(10 \choose 3\) Then we need to place 2 Ls \(7 \choose 2\) Then we need to place 1 H \(5 \choose 1\) Then we need to place 2 Ss \(4 \choose 2\) Then we need to place the two Es (though there is only 2 spots left): \(2 \choose 2\) The number of different ways we can do the whole job is the product of the number of different ways we can do each task in the job. \[{11\choose 1} {10 \choose 3}{7 \choose 2}{5 \choose 1}{4 \choose 2}{2 \choose 2}\]\[=831600\]
i see what i did wrong. thank you so much.
To do it by hand: \[=\frac{11!}{1!(11-1)!}\cdot \frac{10!}{3!(10-3)!}\cdot \frac{7!}{2!(7-2)!}\cdot \frac{5!}{1!(5-1)!}\cdot \frac{4!}{2!(4-2)!} \cdot \frac{2!}{2!(2-2)!}\]\[=\frac{11!}{10!}\cdot \frac{10!}{3!7!}\cdot \frac{7!}{2!5!}\cdot \frac{5!}{1!4!}\cdot \frac{4!}{2!2!} \cdot 1\]\[=\frac{11!}{3!\cdot 2!\cdot 2! \cdot 2!} = \frac{39916800}{48} = 831600\]
That's interesting. I was going to suggest taking 11! and dividing by the extra permuations 3! and (2!)^3
its really nice to see the build up! i know just to go straight to the last line of equations you posted, but ive never seen it explained step by step before.
I agree
gosh. you have been so kind. thank u sooo much.
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