Question

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

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