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Question: Out of \({\text{7}}\) consonants and \({\text{4}}\) vowels, how many words can be made each contain ...

Out of 7{\text{7}} consonants and 4{\text{4}} vowels, how many words can be made each contain 3{\text{3}} consonant and {\text{2 }}vowel?

Explanation

Solution

Hint: - Number of ways of selecting (r r numbers out of nn numbers)  = nCr{\text{ = }}{}^n{C_r}

Number of ways of selecting ( 3{\text{ 3}} consonants out of  7{\text{ 7}}) = 7C3{\text{ = }}{}^7{C_3}
And the number of ways of choosing ( 2{\text{ 2}} vowels out of  4{\text{ 4}}) = 4C2{\text{ = }}{}^4{C_2}
And since each of the first groups can be associated with each of the second,
The number of combined groups, each containing  3{\text{ 3}} consonants and 2{\text{ 2}} vowels, is
7C3×4C2=7!3!(73)!×4!2!(42)!  = 7×6×53×2×1×4×32×1  = 210  \Rightarrow {}^7{C_3} \times {}^4{C_2} = \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}} \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \\\ {\text{ = }}\dfrac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \dfrac{{4 \times 3}}{{2 \times 1}} \\\ {\text{ = 210}} \\\
Number of groups, each having  3{\text{ 3}} consonants and  2{\text{ 2}} vowels  = 210{\text{ = 210}}
Each group contains 5 5 letters
Number of ways of arranging 5 5 letters among themselves  = 5!{\text{ = 5!}}
 = 5×4×3×2×1 =120  {\text{ = }}5 \times 4 \times 3 \times 2 \times 1 \\\ = 120 \\\
  \therefore \; Required number of ways=(210×120)=25200 = (210 \times 120) = 25200.
Hence, the answer is 2520025200.
Note: - Whenever we face such types of questions,we have to first use the method of selection for selecting the numbers of vowels and constants that are given in question, and then by applying the method of rearranging to rearrange the words to get the total number of words.