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?

  • $\begingroup$ Can you explain, why do you think that $C_4^{11}$ is the right answer? $\endgroup$ – Antoine May 4 '16 at 16:32
  • $\begingroup$ What you mean "by using <another word>"? do you mean different factors of length $4$ in the other word? $\endgroup$ – StefanH May 4 '16 at 16:33
  • $\begingroup$ I mean that by using "MISSISSIPPI" can form how many different kind of 4 letter words. $\endgroup$ – firstimer08302 May 4 '16 at 16:34
  • $\begingroup$ Can you repeat letters? $\endgroup$ – N.S.JOHN May 4 '16 at 16:35
  • $\begingroup$ ya repetition is allow, trying to grasp the concept, don't know the formula is used in the right scenario. $\endgroup$ – firstimer08302 May 4 '16 at 16:38

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.

  • $\begingroup$ 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 $4^4$, $\endgroup$ – lulu May 4 '16 at 17:00
  • $\begingroup$ I got $167$ (though mine solution is also prone to error from the same reason that you've mentioned). $\endgroup$ – barak manos May 4 '16 at 17:03
  • $\begingroup$ I note that you don't have $IIII$ for example. $\endgroup$ – lulu May 4 '16 at 17:04
  • 1
    $\begingroup$ 176 is the right answer. $\endgroup$ – Ian Miller May 4 '16 at 17:06
  • $\begingroup$ @IanMiller Thanks. I just worked it with the generating function...I assume that's how you did it? $\endgroup$ – lulu May 4 '16 at 17:10

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