Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

thanks for y'all's help. i'm stumped on another. how many distinct permutations can be formed using the letters in TALLAHASSEE?

OpenStudy (anonymous):

Using all the letters?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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..

OpenStudy (anonymous):

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. \[{11\choose 1} {10 \choose 3}{7 \choose 2}{5 \choose 1}{4 \choose 2}{2 \choose 2}\]\[=831600\]

OpenStudy (anonymous):

i see what i did wrong. thank you so much.

OpenStudy (anonymous):

To do it by hand: \[=\frac{11!}{1!(11-1)!}\cdot \frac{10!}{3!(10-3)!}\cdot \frac{7!}{2!(7-2)!}\cdot \frac{5!}{1!(5-1)!}\cdot \frac{4!}{2!(4-2)!} \cdot \frac{2!}{2!(2-2)!}\]\[=\frac{11!}{10!}\cdot \frac{10!}{3!7!}\cdot \frac{7!}{2!5!}\cdot \frac{5!}{1!4!}\cdot \frac{4!}{2!2!} \cdot 1\]\[=\frac{11!}{3!\cdot 2!\cdot 2! \cdot 2!} = \frac{39916800}{48} = 831600\]

OpenStudy (phi):

That's interesting. I was going to suggest taking 11! and dividing by the extra permuations 3! and (2!)^3

OpenStudy (anonymous):

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.

OpenStudy (phi):

I agree

OpenStudy (anonymous):

gosh. you have been so kind. thank u sooo much.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!