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How many words can be formed from the word 'alpha'? The letter 'a' may be used twice but the other letters may only be used once.

There are no restrictions on whether or not they're real words, just combinations of letters. So, I need to add together:

$5$ letter words $+$ $4$ letter words $+$ $3$ letter words $+$ $2$ letter words $+$ $1$ letter words

There are, $(5!/2!)=60$, 5 letter words. 120 total but then we divide out by duplicate words because we have two a's.

$1$ letter words $=4$

The part I am having trouble with is the duplication of 'a'. I can't seem to figure out how to find $2,3,$ and $4$ letter words. For 2 letter words, there were so few examples I just wrote them out and saw that there were 13. So,

$(5!/3!)-[(2!)3+1]=13$

20 possibilities and then we need to subtract out the ones that contain 2 of p,l, and h, as well as the second aa.

I'm having a hard time believing my work for 2 letter words is accurate and attempting to do 3 and 4 letter words suggests that it isn't. Let me know if there is anything I can to do clear up what is being asked. Thanks in advanced.

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Let us look for example at $4$-letter words. They are of two types: (i) words with at most one $a$ and (ii) words with two $a$'s.

For Type (i), there are $4!$. For Type (ii), we choose the location of the $a$'s in $\binom{4}{2}$ ways. For each such way, the remaining two places can be filled in $(3)(2)$ ways.

The same strategy works for $3$-letter words. There are $(4)(3)(2)$ ways to have at most one $a$. For the words with two $a$'s, the locations can be picked in $\binom{3}{2}$ ways, and for each such way the remaining slot can be filled in $3$ ways.

Note that if we had more non-duplicated letters, the analysis would be much the same.

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  • $\begingroup$ Thanks a lot! Been a long time since I've done these and that helped refresh my memory quite well. Using your logic, does that make 2 letter words: $(4)(3)+1=13$? $\endgroup$ – Vincent Jun 11 '14 at 5:22
  • $\begingroup$ Yes. In the language I used, $(4)(3)$ Type (i) and $1$ Type (ii). $\endgroup$ – André Nicolas Jun 11 '14 at 5:24
  • $\begingroup$ I didn't want to open up a new thread for something that is so similar. I had another problem, there were four parts and I was able to complete the first 3. The last part asks: If 4 married couples buy 8 seats in a row for a concert, how many ways can they be seated if no man sits next to his wife? I tried using a similar approach to what you suggested above. First, choose a couple (4), then choose one of them (2), then fill in the seat next to them (6)=48. Next, I thought to multiply this by 4 (for four couples) but it doesn't seem like that makes any sense. $\endgroup$ – Vincent Jun 11 '14 at 6:14
  • $\begingroup$ The problem has a different flavour. If you post it, someone will probably answer soon. Probably I will not, it is getting late here! $\endgroup$ – André Nicolas Jun 11 '14 at 6:18
  • $\begingroup$ Okay, thanks. Will do. $\endgroup$ – Vincent Jun 11 '14 at 6:21
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Break each case in two parts: (i.e. words with all distinct letters + words with two repeated a's)

No. of 4 letter words formed from 'alpha' = No. of 4 letter words formed with all distinct letters (4C4 * 4!) + No. of 4 letter words formed with two a's already present (3C2 * 4!/2!).

Similarly, No. of 3 letter words = (4C3 * 3!) + (3C1 * 3!/2!)

No. of 2 letter words = (4C2 * 2!) + (3C0 * 2!/2!)

No. of 1 letter words = (4C1 * 1!)


NOTE: a) To form words with distinct letters we choose from ['a', 'l', 'p' and 'h'], hence 4Cx; then we arrange them in x! ways.

b) To form words with two a's already present, we need to choose from ['l', 'p' and 'h'], hence 3Cx; and arrange them in x!/2! ways.

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