There are 9 letters in "telephone". We are using permutations instead of combinations because order matters ("tele" and "elet" are not the same thing).

Total permutations of 4 letters from 9:

9P4 = 9!/5! = 3024

However, note that "e" appears 3 times. This will lead to many duplicates. So we need to actually count how many perms of 4 letters there are using the 3 e's and the other 6 letters separately. We can then add them together to get the total number of words.

[words with 0 e] + [words with 1 e] + [words with 2 e] + [words with 3 e]

[3P0 * 6P4] + [3P1 * 6P3] + [3P2 * 6P2] + [3P3 * 6P1]

[1 * 360] + [3* 120] + [6 * 15] + [6 * 6]

360 + 360 + 90 + 36

846

Edit: OK I better try this again before my tutoring license gets revoked. So I think my mistake here was that the e's are actually combinations not perms because the order does not matter for them (e1-e2 = e2-e1, but t-l and l-t are not the same). I then forgot to add to the total positions of the non-e's (I had the different orders but forgot those orders can be in multiple spots), which once again is a combination. So my answer uses both perms and combos, and I wound up with:

[3C0 * 6P4 * 4C4] + [3C1 * 6P3 * 4C3] + [3C2 * 6P2 * 4C2] + [3C3 * 6P1 * 4C1]

[1 * 360 * 1] + [3* 120 * 4] + [3 * 15 * 6] + [1 * 6 * 4]

360 + 1440 + 270 + 24

2094

This looks to be consistent with what Stanton was describing, but if this is still incorrect, then I'd invite any of the other tutors to actually give this problem a shot.

Mark M.

_{n}P_{r}) / x_{1}!x_{2}! where x is the number of repititions._{9}P_{4}/ 3!02/22/17