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We have two vowels $a,e$ and four consonants $b,c,d,f$. To find the number of three-letter words each containing one vowel at least, we will consider the following cases:

The case I: Exactly one vowel and two consonants.

In this case, one vowel out of two vowels can be selected in ${}^2{C_1} = 2$ ways. Two consonants out of four consonants can be selected in ${}^4{C_2} = \dfrac{{4 \times 3}}{{1 \times 2}} = \dfrac{{12}}{2} = 6$ ways.

All three letters (one vowel and two consonants) can arrange among themselves in $3! = 6$ ways. Therefore, the total number of three-letter words each containing exactly one vowel are $2 \times 6 \times 6 = 72$.

Case II: Exactly two vowels and one consonant.

Two vowels out of two vowels can be selected in ${}^2{C_2} = \dfrac{{2 \times 1}}{{1 \times 2}} = 1$ way. One consonant out of four consonants can be selected in ${}^4{C_1} = 4$ ways.

All three letters (one vowel and two consonants) can arrange among themselves in $3! = 6$ ways. Therefore, the total number of three-letter words each containing exactly two vowels is $1 \times 4 \times 6 = 24$. Now we will add the possibilities of the case I and case II to get the required number of words. Therefore, the total number of three-letter words each containing at least one vowel is $72 + 24 = 96$.

If we consider the order of objects then it is a permutation. If we do not consider the order of objects then it is a combination. ${}^n{C_r}$ gives the total number of ways of selecting $r$ objects out of $n$ objects. ${}^n{P_r}$ gives the total number of distinct arrangements when we arrange $r$ objects among $n$ objects. The number of permutations of $n$ distinct objects is $n!$.

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