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How many 4 letter words can be formed by the word "mississippi" so we have letter "i" in all arrangements?

What I've done was to include "i" to one those 4 words, so we would have 3 words to choose then considering the number of similar letters.

Is this correct?

$$\frac{3!}{4!2!}\times 4!$$

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    $\begingroup$ Hard to parse your formula, but your reasoning sounds wrong. The problem is that your method leads to overcounting...as you count words with more than one $i$ more than once. $\endgroup$ – lulu Oct 13 '18 at 12:09
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    $\begingroup$ Easier to first ignore the condition on the $i$, then subtract off the words that do not contain an $i$. $\endgroup$ – lulu Oct 13 '18 at 12:10
  • $\begingroup$ this question should help to get you started. $\endgroup$ – lulu Oct 13 '18 at 12:13
  • $\begingroup$ Look at this question $\endgroup$ – rogerl Oct 13 '18 at 15:16
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We wish to count how many four-letter words can be formed from the letters of the word MISSISSIPPI that include at least one I.

The word MISSISSIPPI has $1$ M, $4$ Is, $4$ Ss, and $2$ Ps.

We consider cases, depending on how many Is are included.

Case 1: Exactly one I is used. There are three subcases.

  1. Exactly one I is used and another letter is used three times.
  2. Exactly one I is used, another letter is used twice, and a different letter is used once.
  3. Exactly one I is used and three different letters are each used once.

Exactly one I is used and another letter is used three times: The only letter other than I that can be used three times is S. There are $\binom{4}{3}$ ways to choose which positions are occupied by the three Ss. The position of the I is then fixed. Hence, there are $$\binom{4}{3}$$ admissible arrangements in this case.

Exactly one I is used, another letter is used twice, and a different letter is used once: The letter that is used twice must be a P or an S. That gives us two choices. Once we choose that letter, we are left with two choices for the letter that is used once (M and the letter we did not choose). There are $\binom{4}{2}$ ways to choose the positions of the letter that is used twice and $2!$ ways to arrange the remaining letters in the remaining two positions. Hence, there are $$\binom{2}{1}\binom{2}{1}\binom{4}{2}2!$$ admissible arrangements in this case.

Exactly one I is used and three different letters are each used once: Then four different letters are each used once. They can be arranged in $4!$ ways.

Case 2: Exactly two Is are used: There are two subcases.

  1. Exactly two Is are used and another letter is also used twice.
  2. Exactly two Is are used and two different letters are each used once.

Exactly two Is are used and another letter is also used twice: The other letter must be an S or a P since only one M is available. Thus, we have two choices for the other letter that appears twice and $\binom{4}{2}$ ways to choose which two positions are occupied by that letter. The remaining positions must be filled with Is. Hence, there are $$\binom{2}{1}\binom{4}{2}$$ admissible arrangements in this case.

Exactly two Is are used and two different letters are each used once: There are $\binom{3}{2}$ ways to select which two of the other three letters are each used once. There are $\binom{4}{2}$ ways to select the positions of the two Is and $2!$ ways to arrange the remaining letters in the remaining two positions. Hence, there are $$\binom{3}{2}\binom{4}{2}2!$$ admissible arrangements in this case.

Case 3: Exactly three Is are used: There are three ways to choose which of the other letters will be used in the word and four ways to choose the position of that letter. The remaining positions must be filled with Is. Hence, there are $$\binom{3}{1}\binom{4}{1}$$ admissible arrangements in this case.

Case 4: Four Is are included.

There is only one way to do this since we must fill all four positions with an I.

Total Since these cases are mutually exclusive and exhaustive, the number of four-letter words that can be formed from the letters of the word MISSISSIPPI is $$\binom{4}{3} + \binom{2}{1}\binom{2}{1}\binom{4}{2}2! + 4! + \binom{2}{1}\binom{4}{2} + \binom{3}{2}\binom{4}{2}2! + \binom{3}{1}\binom{4}{1} + \binom{4}{4}$$

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  • $\begingroup$ What i didn't understand is in case1 and sub-case 2 why did you use 1 from 2 twice? is that for P and S? and if we had another letter for example 1 C the answer would have not been changed right? $\endgroup$ – lighting Oct 14 '18 at 19:49
  • $\begingroup$ We have to choose which letter is used twice. It must be P or S, so we must choose one of those two. Say we select P. Then we have two choices for the letter other than I that is used once since it must be M or S. $\endgroup$ – N. F. Taussig Oct 14 '18 at 19:52
  • $\begingroup$ Thanks! if instead of four M, I, S and P letters we had 5 letters (for example we had another letter like 1 L) all answers except of case1 sub-case 1 and case2 sub-case1 would have been changed? $\endgroup$ – lighting Oct 14 '18 at 20:15
  • $\begingroup$ Clearly, case 4 would not change since there is only one way to place four Is in four positions. Otherwise, what you said is correct. $\endgroup$ – N. F. Taussig Oct 14 '18 at 20:23
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Okay, so just breaking your question down, a simple way to do any question like this is to just, list out the ways in which you can create $4$ letter words.
We can do it in $4$ ways in this case. You can have $4$ distinct letters, or you can have $2$ same letters and $2$ different, or you can have $2$ of the same letter and another $2$ of the same or all $4$ same letters.
Now,
Case $1$: All four distinct. In this case, you can have only $4$ letters from $4$ so you choose $4$ and then arrange. Which gives us, $$\binom {4}{4} \cdot 4!$$ Case $2$: Two same and two are same. That gives us, $2$ letters to choose from, and then arranging them gives us, $$\binom {2}{1} \cdot \dfrac{4!}{2!\cdot2!}$$
Case $3$: Two same and two are distinct. Now, in this case, you can have $2$ cases again, wherein you choose the $i^{'}$s in the double category or a distinct one. If you choose the $i^{'}$s to appear in doublets then you will have $$\binom {3}{1} \cdot \dfrac{4!}{2!}$$ and when the $i^{'}$s are distinct then you have $$\binom{2}{1} \cdot \binom{2}{2} \cdot \dfrac{4!}{2!}$$ Case $4$: All are same. In this case we have only $1$ case which is to pick all $4$ $i^{'}$s.
Thus your final count is, $$\sum^4 _{i=1} C_i = \binom {4}{4} \cdot 4! + \binom {2}{1} \cdot \dfrac{4!}{2!\cdot2!} + \binom {3}{1} \cdot \dfrac{4!}{2!} + \binom{2}{1} \cdot \binom{2}{2} \cdot \dfrac{4!}{2!} + 1$$

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  • $\begingroup$ You are missing some cases. $\endgroup$ – N. F. Taussig Oct 14 '18 at 15:32
  • $\begingroup$ yes, I submited the answer in a hurry $\endgroup$ – Prakhar Nagpal Oct 15 '18 at 15:37

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