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As the question states, how many four letter words can be made from the multi-set $\left\{T,E,L,E,P,H,O,N,E\right\}$? One condition applies, in that (EELE) is a valid four letter word but (EEEE) is not. I was thinking somewhere along the lines of placing k balls into n boxes, the general example for counting, where the k balls are the letters and boxes are the four possible places in our four letter word, but I just can't seem to form a believable answer.

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  • $\begingroup$ You could separate out cases with 1E, 2Es, and 3Es. All other letters are distinct so counting words in each case should be straighforward. $\endgroup$
    – user96614
    Nov 6, 2013 at 17:41

2 Answers 2

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From the word $$\left\{T,L,P,H,O,N,E,E,E\right\}$$ we count number of four leter words with no E $\binom{6}{4}4!$, with one E $\binom{6}{3}4!$, with two E $\binom{6}{2}\frac{4!}{2!}$ and with three E $\binom{6}{1}\frac{4!}{3!}$ and get
$$\binom{6}{4}4!+\binom{6}{3}4!+\binom{6}{2}\frac{4!}{2!}+\binom{6}{1}\frac{4!}{3!}=$$

$$=15\cdot24+20\cdot24+15\cdot12+6\cdot4$$

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You have $9$ letters and a four letter word. So, it needs a permutation:

$9*8*7*6 = 3024$

That is, each of the $9$ letters can be combined with the remaining $8$ letters, that can be combined with the remaining $7$ letters, that can be combined with the remaining $6$ letters.

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    $\begingroup$ A permutation will not work since there are some repeated elements in the set. $\endgroup$
    – nispio
    Nov 6, 2013 at 18:56

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