# How many combinations are there of the word “clementine” with a maximum with 3 repeated consonants?"

I'm having an issue with a math question:

"How many combinations are there of the word "clementine" with a maximum with 3 repeated consonants?"

The issue I have is understanding how to account for the combinations of the consonants.

As there are 4 vowels and 6 consonants but at most only 3 consonants should be next to each other, so...

(1 C) (1 L) (2 N) (1 M) (1 T) <- Consonants

(3 E) (1 I) <- Vowels

So far I believe I'm on the correct path however I am confused on calculating how many spaces the 3 letters can fit into:

## 4!/(2!1) X 6!/(3!1)X (the number of spaces that 3 consonants can group in?)

Any help in understanding the process for this question better would be appreciated. I'm aware there are many questions on the subject on stackoverflow and have looked at a number including :

How many permutations of a word do not contain consecutive vowels?

However, I'm uncertain how to handle the specific instance of having multiple consonants together in the space.

• When you say "repeated consonants", do you mean "consecutive consonants"? – Joffan Mar 20 '17 at 22:44
• yes that is what i meant, sorry for the confusing wording – D3181 Mar 20 '17 at 23:55

## 2 Answers

We will count the ways we can assign positions to the consonants, then count the number of ways the consonants and vowels can be assigned to their designated positions.

An arrangement of the word CLEMENTINE in which no more than three consonants are consecutive has the form $$\square v \square v \square v \square v \square$$ where $v$ represents the position of a vowel and the squares represent the possible positions of the consonants. Let $x_k$ represent the number of consonants placed in the $k$th square from the left. Since there are six consonants and no more than three consonants can be placed in each position represented by a square, $$x_1 + x_2 + x_3 + x_4 + x_5 = 6 \tag{1}$$ where $x_k \leq 3, 1 \leq k \leq 5$. Equation 1 is an equation in the non-negative integers. A particular solution corresponds to the placement of four addition signs in a row of six ones. For instance, $$1 + 1 + + 1 1 1 + 1$$ represents the solution $x_1 = x_2 = 1$, $x_3 = 0$, $x_4 = 3$, and $x_5 = 1$, while $$1 + 1 + 1 + 1 + 1 1$$ represents the solution $x_1 = x_2 = x_3 = x_4 = 1$, and $x_5 = 2$. Therefore, the number of solutions of equation 1 in the nonnegative integers is the number of ways four addition signs can be placed in a row of six ones, which is $$\binom{6 + 4}{4} = \binom{10}{4}$$ since we must choose which four of the ten positions (six addition signs and four addition signs) will be filled with addition signs.

However, we have counted solutions of equation 1 in which one or more of the variables is larger than $3$. We must exclude these. Note that at most one variable can exceed $3$ since $2 \cdot 4 = 8 > 6$.

Suppose $x_1 > 3$. Since $x_1$ is an integer, $x_1 \geq 4 \implies y_1 = x_1 - 4$ is a nonnegative integer. Substituting $y_1 + 4$ for $x_1$ in equation 1 yields \begin{align*} y_1 + 4 + x_2 + x_3 + x_4 + x_5 & = 6\\ y_1 + x_2 + x_3 + x_4 + x_5 & = 2 \tag{2} \end{align*} Equation 2 is an equation in the nonnegative integers. The number of solutions of equation 2 is the number of ways four addition signs can be placed in a row of two ones, which is $$\binom{2 + 4}{4} = \binom{6}{4}$$ By symmetry, this is the number of solutions in which $x_k > 3$, $1 \leq k \leq 5$. Hence, the number of solutions of equation 1 in which none of the variables exceeds $3$ is $$\binom{10}{4} - \binom{5}{1}\binom{6}{4}$$ Thus far, we have counted the number of ways we can designate positions for the consonants. Once we have chosen the positions where the consonants will be located (which determines the positions where the vowels will be located), we must arrange the consonants and vowels in those positions. There are six consonants, of which there are $2$ N's, $1$ C, $1$ L, $1$ M, and $1$ T. We choose two of the six positions for the $N$'s, then can place the remaining four consonants in the four remaining positions designated for the consonants. This can be done in $$\binom{6}{2}4!$$ ways. That leaves the four positions for the four vowels, of which there are $3$ E's and $1$ I. Choosing which of the four positions will be assigned an I completely determines how the vowels will be arranged in their designated positions. Hence, the number of possible arrangements of the letters of the word CLEMENTINE in which no more than three vowels are consecutive is $$\left[\binom{10}{4} - \binom{5}{1}\binom{6}{4}\right] \cdot \binom{6}{2}4! \cdot 4$$

It may be simplest to find the number of words with exactly six consecutive consonants, then exactly five consecutive consonants, then exactly four consecutive consonants, add them up, and subtract this total from the total number of words possible.

The total number of configurations possible (without constraints) is $10!/(3! 2!) = 302,400$, where the denominator is for the fact that there are three $e$s and two $n$s.

six consecutive consonants

Of the ten "slots" for letters, if there are exactly six consecutive consonants, the first one must appear in slot 1 or slot 2 or slot 3 or slot 4 or slot 5. That's all. For each such "starting slot", you have $c$, $l$, $m$, $n$ and $t$ (with $n$ appearing twice). The number of distinct ways to fill in those six slots is then $6!/2!$. For each of these, you have the four vowel slots for the 3 $e$ and one $i$. There are only four places to put the $i$, and the remaining $e$s are "forced."

five consecutive consonants

If there are exactly five consecutive consonants, the first "slot" must be slot 1 or slot 2 or slot 3 or slot 4 or slot 5 or slot 6. For each of these, however, there is a different number of places where the sixth consonant can be placed. As above, once the consonants have been placed, there are only four possible words, corresponding to the four possible positions for the $i$.

four consecutive consonants

The same logic as above holds, but is a bit more complex.

Hope this helps!