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Question: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?...

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Explanation

Solution

First of all, find the number of ways of choosing the vowels and consonants separately. Then, find the ways to choose both of them. And finally, find the number of ways of arranging them by multiplying it with 5!.

Complete step-by-step answer:
In the question, we are given that out of 7 consonants and 4 vowels, we need to find how many 5 letters words consisting of 3 consonants and 2 vowels can be formed. So, first of all, we need to know what Combination in Mathematics is. In Mathematics, the combination is a solution of items from a collection, such that the order doesn’t matter. For example, we are given 3 fruits, say an apple, an orange, and a pear. 3 combinations can be drawn from this set: an apple, a pear; an orange, a pear; an apple, an orange. More formally, a K – combination of the set S is a subset of the distinct elements of S. If a set has ‘n’ elements, the number of K – combinations is equal to binomial coefficient,
nCk=n(n1)(n2).....(nk+1)k(k1)(k2).....1^{n}{{C}_{k}}=\dfrac{n\left( n-1 \right)\left( n-2 \right).....\left( n-k+1 \right)}{k\left( k-1 \right)\left( k-2 \right).....1}
which can also be written using factorials as n!k!(nk)!\dfrac{n!}{k!\left( n-k \right)!} where knk\le n and is 0 when k > n. The set of all k – combinations of a set often denoted by sCk^{s}{{C}_{k}}.
Now in the question, we need to choose 3 consonants and 2 vowels from the given 7 consonants and 4 vowels. The number of ways in which consonants can be chosen is
7C3=7!4!×3!=35^{7}{{C}_{3}}=\dfrac{7!}{4!\times 3!}=35
Now for finding the number of ways in which 2 vowels can be selected from 4 vowels is
4C2=4!2!×2!=6^{4}{{C}_{2}}=\dfrac{4!}{2!\times 2!}=6
So, the number of ways of selecting only consonants is 35 and vowels is 6. Hence, for selecting both consonants and vowels is:
35×6=210 ways\Rightarrow 35\times 6=210\text{ ways}
So now, we need to form the word by 5 distinct letters, so that can be arranged in 5! ways which is equal to 120 ways. So, the total number of words that can be formed is
120×210=25200120\times 210=25200
Hence, there are a total of 25200 words that can be formed.

Note: Students generally get confused between permutations and combinations as combination means only choosing while permutation means first choosing and then arranging. First we find 5 letters out of given letters and then arrange them using the formula 5! = 5×4×3×2×1 5 \times 4 \times 3 \times 2 \times 1.