Find the number of words each consisting of $3$ consonants and $3$ vowels that can be formed from the letters of the word CIRCUMFERENCE.In how many of these $c$'s will be together?
The consonants are $c,r,c,m,f,r,n,c$ and the vowels are $i,u,e,e,e$.
Number of ways of arranging 3 vowels$=\frac{3!}{2!}+1+3!=10$
I am stuck in counting the number of ways of arranging 3 consonants because $c$ is coming 3 times and $r$ is coming 2 times.
Best Answer
Arrangements of the vowels:
You correctly calculated that the number of distinguishable ways to arrange three E's is $1$ and that the number of ways to arrange three different vowels is $3!$. However, you calculated the number of arrangements with two E's incorrectly.
There are $\binom{3}{2}$ ways of selecting the positions of the two $E$'s and two ways of selecting the other vowel. Hence, the number of arrangements with two E's is $$\binom{3}{2}\binom{2}{1}$$ Hence, the number of ways of arranging the three vowels in the three positions selected for the vowels is $$3! + \binom{3}{2}\binom{2}{1} + 1$$
Arrangements of the consonants:
Case 1: Three different consonants.
We can choose three of the five consonants, then arrange them in order in $$\binom{5}{3} \cdot 3!$$ ways.
Case 2: Two different consonants.
We choose which of the two letters C or R is repeated, choose two of the three positions in which to place the repeated letter, then choose one of the other four letters to fill the open slot, which yields $$\binom{2}{1}\binom{3}{2}\binom{4}{1}$$ possible arrangements.
Case 3: One consonant is used.
There is only one way to fill the three positions with a C.
Hence, the number of ways of arranging the consonants within their three selected positions is $$\binom{5}{3}\cdot 3! + \binom{2}{1}\binom{3}{2}\binom{4}{1} + 1$$
Number of words consisting of three consonants and three vowels from the letters of the word CIRCUMFERENCE
We choose three of the six positions for the vowels, arrange the vowels in the selected locations, then arrange the consonants in the three remaining positions, which yields $$\binom{6}{3}\left[3! + \binom{3}{2}\binom{2}{1} + 1\right]\left[\binom{5}{3}\cdot 3! + \binom{2}{1}\binom{3}{2}\binom{4}{1} + 1\right]$$