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How many different $4$ letter words can be formed by using "MISSISSIPPI".

my answer to this will be $\binom{11}{4}=330$.

Is this correct?

JMoravitz
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firstimer08302
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  • Can you explain, why do you think that $C_4^{11}$ is the right answer? – Antoine May 04 '16 at 16:32
  • What you mean "by using "? do you mean different factors of length $4$ in the other word? – StefanH May 04 '16 at 16:33
  • I mean that by using "MISSISSIPPI" can form how many different kind of 4 letter words. – firstimer08302 May 04 '16 at 16:34
  • Can you repeat letters? – N.S.JOHN May 04 '16 at 16:35
  • ya repetition is allow, trying to grasp the concept, don't know the formula is used in the right scenario. – firstimer08302 May 04 '16 at 16:38
  • To clarify: is $MMMM$ a possible word on your list? Presumably $SSSS$ is possible (as there are four $S's$ to choose from). But there is only the one $M$ so I'd have assumed that you couldn't make $MMMM$. – lulu May 04 '16 at 16:43
  • Recommend breaking into cases. How many words can be made if all letters are distinct? How many words can be made if exactly one letter occurs twice (e.g. miss)? if exactly one letter occurs thrice (e.g. siss)? if two letters are repeated (e.g. piip)? if a letter occurs four times (e.g. ssss). The number $\binom{11}{4}$ unfortunately is incorrect for this situation. Remember that order of letters within words matters and that not all of your eleven letters are distinct. – JMoravitz May 04 '16 at 16:48
  • An answer of $\binom{11}{4}$ would count the number of subsets (where order doesn't matter) of letters from an eleven letter word where all eleven letters are different. An answer of $\binom{11}{4}4!$ would count the number of four letter words made from letters from an eleven letter word where all eleven letters are different. Consider the question of "how many four letter words can you make from the word AAAAAAAAAAA?" By your logic, the answer would still have been $\binom{11}{4}$, but there is clearly only one possible: AAAA. – JMoravitz May 04 '16 at 16:51
  • Duplicate of http://math.stackexchange.com/questions/20238/6-letter-permutations-in-mississippi and http://math.stackexchange.com/questions/517101/5-letter-arrrangments-of-mississippi The first one covers a technique which will work for any length of word. – Ian Miller May 04 '16 at 17:02
  • Number of words of length $n$ is given by: ${1, 4, 15, 53, 176, 550, 1610, 4340, 10430, 21420, 34650, 34650}$ – Ian Miller May 04 '16 at 17:05

1 Answers1

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To clarify: I am assuming that this works like Scrabble. That is, you have exactly those $11$ letters from which to choose. Thus, in particular, $MMMM$ would not be allowed. If, to the contrary, you can reuse letters then the answer is trivially $4^4$.

I think the easiest way to count these is to track the $M's$ and $P's$. After all, these are the only constrained ones (as both $SSSS$ and $IIII$ are possible). Accordingly, let $(m,p)$ denote the case in which exactly $m$ $M's$ and $p$ $P's$ appear. Of course $m\in \{0,1\}$ and $p\in \{0,1,2\}$. We remark that once the $M's$ and $P's$ are settled, you can choose between $S,I$ freely for the other slots. We work the six cases separately.

$(0,0)$. We have four slots to fill however we like with $S,I$ so $2^4=\fbox {16}$

$(1,0)$. We have four ways to place the $M$ and then three free slots so $4\times 2^3=\fbox {32}$

$(0,1)$. As in the case $(0,1)$ we get $\fbox {32}$

$(0,2)$. we have $\binom 42 = 6$ ways to place the $P's$ and then two free slots so $6\times 2^2=\fbox {24}$

$(1,1)$. Four ways to place the $M$, then three ways to place the $P$, and then two free slots so $4\times 3\times 2^2=\fbox {48}$

$(1,2)$. Four ways to place the $M$ then three ways to place the two $P's$ and then one free slot so $4\times 3\times 2=\fbox {24}$

FInally we get $$16+32+32+24+48+24=\fbox {176}$$

Note: while I wouldn't say the preceding calculation was difficult it is certainly error prone so I advise checking it carefully.

lulu
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  • Yes, but as I asked in the comments above, my guess is that the OP was talking about repeating the $S's$ and $I's$. That is, I do not believe that $MMMM$ is allowed. Of course the OP should clarify. If things like $MMMM$ are allowed then the count is trivially $4^4$, – lulu May 04 '16 at 17:00
  • I got $167$ (though mine solution is also prone to error from the same reason that you've mentioned). – barak manos May 04 '16 at 17:03
  • I note that you don't have $IIII$ for example. – lulu May 04 '16 at 17:04
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    176 is the right answer. – Ian Miller May 04 '16 at 17:06
  • @IanMiller Thanks. I just worked it with the generating function...I assume that's how you did it? – lulu May 04 '16 at 17:10
  • Yeh, I used the generating function. I wrote some Mathematica code a while back where I can enter any word and it calculates the generating function and pulls out all the coefficients so this was a very easy one to check. – Ian Miller May 04 '16 at 17:11
  • Actually even if repetition is allowed $4^4$ will not be the answer, because the formula takes repeated letter words as different. The formula sees $ MMSI$ as two words instead of one(as M repeats) – N.S.JOHN May 04 '16 at 17:21
  • @N.S.JOHN Not following. What formula? To me "with repetition" just means that each slot can have $M,S,I,P$ with no restrictions. Hence $4^4$. – lulu May 04 '16 at 17:27
  • $4^4$ is only true if we see $MMSI$ and its kind as twodifferent words instead of one. – N.S.JOHN May 04 '16 at 17:29
  • @N.S.JOHN Still not following. $MMSI$ is clearly just one word. With repetition there are $4$ choices for the first slot, $4$ for the second slot, and so on. Hence $4\times 4\times 4 \times 4 = 4^4$. – lulu May 04 '16 at 17:35
  • Yes you are right I made a mistake thanks. – N.S.JOHN May 04 '16 at 17:38