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"How many ways are there to arrange the letters in the word 'MISSISSIPPI' in such a way that there are no three consonants in a row?"

I am thinking like this. The following are 'slots' for the letters of our word: _ _ _ _ _ _ _ _ _ _ _. There are 21 consonants in total (English). The first slot as 21 possibilities, the second also has 21, but the third has only 5 possibilities. So how many ways can I place vowels within this word... I don't know.

I can try to enumerate all of the place the I's can go, but there has to be a better way than doing that.

Ozera
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4 Answers4

7

MISSISSIPPI has seven consonants and four vowels (which all happen to be I). Write X for any consonant, and first think of how many possible patterns of the type IXXIXIXXIXX, with seven Xes and four Is exist, with no three Xs in a row. Then think about how to assign the letters MSSSSPP to the Xs. Note that the two subproblems are quite similar to each other. The final step is a simple multiply.

Edit: A bit more on the first subproblem. You'll have to do a bit of hand counting here, I think. There are only four Is, so consecutive Xs can form at most five groups. But there must be at least four groups of Xs, since no group can have more than two Xs. So the configuration of Xs is either XX XX X X X or permutation thereof, of XX XX XX X or permutations thereof. It should be easy to count the permutations in each case. In the first, case, the Is must go one between each group, but in the second, three Is are needed to fill the gaps, and the fourth I can be placed next to one of the other three, or at the beginning or end – five possibilities there.

Harald Hanche-Olsen
  • 31,960
  • So the first subproblem, we have to figure out how many different ways we can make a pattern with no $3$ X's in a row. This is what I tried to ask, but didn't/don't know how to do. – Ozera Jul 05 '14 at 15:56
  • You should have made that clear. Rule #1: Always explain your work and tell where you got stuck. I added some detail on that subproblem. The rest is elbow grease, which I insist you provide yourself. – Harald Hanche-Olsen Jul 05 '14 at 16:11
  • Is the answer 3150? – Ozera Jul 05 '14 at 18:59
  • Yes, 3150. (I said 3675 earlier, but I found a simple mistake in my calculation.) – Harald Hanche-Olsen Jul 05 '14 at 21:37
  • Yay! Thank you for taking your time for the assistance. I understand it all now. – Ozera Jul 05 '14 at 23:59
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The number of solutions to $x_1+x_2+x_3+x_4+x_5 = 7$ with $0 \leq x_i \leq 2$ is same as the coefficient of $x^7$ in $(1+x+x^2)^5$. This can be expanded using binomial theorem as $(1+x(1+x))^5$. Since $x^7$ will occur only in the terms $\binom{5}{4}x^4(1+x)^4$ and $\binom{5}{5}x^5(1+x)^5$, the required coefficient is $\binom{5}{4}\binom{4}{3} + \binom{5}{5}\binom{5}{2} = 30$. Since there are $\binom{7}{4,2,1} = 105$ ways to place $SSSSPPM$ in these 7 places, the required answer is 3150.

Muralidharan
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One way you could do this is first arrange the four I's in a row, creating 5 gaps in which the consonants can be placed.

If we let $x_i$ be the number of consonants in gap $i$, then $x_1+\cdots +x_5=7$ where $0\le x_i\le2$ for each $i$.

If you calculate the number of solutions to this equation, and then multiply by the number of ways to arrange the consonants SSSSPPM in order, I believe you will get the answer you obtained.

user84413
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2

The generating function for the given condition is:

\begin{align*} G(v,c) &= \frac{1+c+c^2}{1-v\left(1+c+c^2\right)} \end{align*}

Since there is one vowel $i$ and three consonants $m,s,p$, we may substitute in the gf to get:

\begin{align*} G(m,s,p,i) &= \frac{1+(m+s+p)+(m+s+p)^2}{1-i\left(1+(m+s+p)+(m+s+p)^2\right)} \end{align*}

and we need to find $[ms^4p^2i^4]$

If the number of M, S, P, I are ${\rm m,s,p,i}$ respectively, then, from the gf:

\begin{align*} \mathrm{N}{\rm(m,s,p,i)} &= \binom{\mathrm{m}+\mathrm{p}+\mathrm{s}}{\mathrm{m},\mathrm{p},\mathrm{s}} \sum_{j=0}^{\mathrm{i}+1}\binom{\mathrm{i}+1}{j}\binom{j}{\mathrm{m}+\mathrm{s}+\mathrm{p}-j} \end{align*}

For ``MISSISSIPPI'', \begin{align*} \mathrm{N}(1,4,2,4) &= 3150 \end{align*}

gar
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