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In how many ways can a consonant and a vowel be chosen out of the letters of the word courage?

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Hint – Pick out the vowels and consonants separately from the given word ‘COURAGE’ and use the combination formula for selecting r dissimilar things out of total n things
Now in the given word ‘COURAGE’ there are 3 consonants that are (c, r, g) and there are 4 vowels that are (o, u, a, e).
So out of 3 consonants a single consonant can be chosen in $^3{C_1}$ways.
Now \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\] using this we can solve $^3{C_1}$ which will be \[\dfrac{{3!}}{{1!(3 - 1)!}} = \dfrac{{3!}}{{2!}} = \dfrac{{3 \times 2!}}{{2!}} = 3\]
So out of 4 vowels a single vowel can be chosen in $^4{C_1}$ways.
Again using the similar concept as above we can solve for $^4{C_1}$ which will be \[\dfrac{{4!}}{{1!(4 - 1)!}} = \dfrac{{4!}}{{3!}} = \dfrac{{4 \times 3!}}{{3!}} = 4\]
Hence the total number of ways of choosing a vowel and a consonant will be $4 \times 3 = 12$ ways

Note- In English alphabet set we have in total 27 alphabets which can broadly be classified into two categories that are vowels and consonants. (A, E, I, O, U) are the vowels while rest are marked as consonants.