Question

Here is a challenging counting / combinatorics problem:

How many 5-letter permutations are there of the word "MATHEMATICS" ? Explain your solution. :)

Answer 1

guess the answer is 11 ! /8 assuming you cannot replace letters

Answer 2

well that is silly, not 11! /8 but rather
\[\frac{11\times 10\times 9\times 8\times 7}{8}\]

Answer 3

interpreting the question to mean essentially you have 11 tiles with those letters on them like in scrabble.

Answer 4

wouldn't it be 11*10*9*8*7?

Answer 5

well it would be but you cannot tell the two m's apart
in your counting mathm counts twice

Answer 6

likewise for A and T which is you divide by 8

Answer 7

what about the A and T's =- oh i see

Answer 8

2 * 2 * 2

Answer 9

yeah that explains the 8 in the denominator. if you want to write something really stupid you could put
\[\frac{11!}{(11-5)!2!2!2!}\] and make it look fancy

Answer 10

right

Answer 11

really?

Answer 12

i beleive you it is early

Answer 13

@Dangol -- no, sorry it is is not that. Your answer is also too large.

Answer 14

i got 6930 is that wrong?

Answer 15

HINT:
There are three cases we must consider, with respect to repetitions.

1) No repeated letters
2) 1 pair of repeated letters
3) 2 pairs of repeated letters

Answer 16

Sorry, Satellite, your answers are too SMALL., my mistake. :(

Answer 17

The answer is between 55,440 and 6,930. I double-checked.

Answer 18

give me a minute see if i can find my mistake. "minute" means ...

Answer 19

ok i still don't see it so let me tell you what i was thinking and you can tell me why i am wrong.

oh wait i think i see the problem.

Answer 20

i shouldn't divide the whole thing by 8. only divide the ones that have 2 m's, 2 a's or to t's by 2
yikes this is going to take me forever

Answer 21

oh and then of course there is the possibility that it has 2 a's and 2 t's
etc

Answer 22

give me something to think about while i go run my saturday errands. don't tell me the answer, but is there a snap way of doing this of doing this?

Answer 23

You have to take the three cases individually and add them:

1) How many 5-letter permutations of MATHEMATICS are there, where there are no repetitions.

2) How many 5-letter permutations of MATHEMATICS are there, where there are one pair of repeated letters?

3) How many 5-letter permutations of MATHEMATICS are there, where there are two pairs of repeated letters?

Part 1 is the easiest to answer.

Answer 24

got it!
i mean i have the idea, not the answer. thnx

Answer 25

a quick calculation gives me 13560