The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Answer
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Hint: Before attempting this question, one should have prior knowledge about the concept of permutation and combination and also remember to use the formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$and $^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, use this information to approach the solution.
Complete step-by-step answer:
According to the given information we have to choose 2 different vowels from 5 vowels and 2 different from 21 constants
So, to choose 2 vowels from 5 vowels the number of ways is $^5{C_2}$
We know that formula of combination is given as $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$here n is the total numbers of objects in the given set and r is the number of objects we have to choose from the set
Therefore, $^5{C_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!2!}} = \dfrac{{5 \times 4}}{2}$
$^5{C_2}$= 10 ways
Also, we have to choose 2 different consonants from 21 constants so the number of possible ways is $^{21}{C_2}$
Therefore, $^{21}{C_2} = \dfrac{{21!}}{{\left( {21 - 2} \right)!2!}} = \dfrac{{21 \times 20}}{2}$
$^{21}{C_2}$= 210 ways
So, the number of possible arrangements of 4 letters is $^4{P_4} = \dfrac{{4!}}{{\left( {4 - 4} \right)!}} = 4 \times 3 \times 2 \times 1 = 24$
So, the total number of ways to select the 4 letters = 24 ways
Thus, the total numbers of words that can be formed is equal to the number of ways selecting 2 vowels $ \times $ number of ways selecting 2 consonants $ \times $ number of arrangements of 4 letters
Therefore, the total number of words of 4 letters = $^{21}{C_2}{ \times ^5}{C_2} \times 4! = 210 \times 10 \times 24 = 50400$
Therefore, the total number of words of 4 letters is 50400 words.
Note: In the above solution we used the method of permutation and combination to get the required result where permutation can be explained as arrangement of an element from the set into a sequence whereas the combination defines the ways of choosing the elements from the set.
Complete step-by-step answer:
According to the given information we have to choose 2 different vowels from 5 vowels and 2 different from 21 constants
So, to choose 2 vowels from 5 vowels the number of ways is $^5{C_2}$
We know that formula of combination is given as $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$here n is the total numbers of objects in the given set and r is the number of objects we have to choose from the set
Therefore, $^5{C_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!2!}} = \dfrac{{5 \times 4}}{2}$
$^5{C_2}$= 10 ways
Also, we have to choose 2 different consonants from 21 constants so the number of possible ways is $^{21}{C_2}$
Therefore, $^{21}{C_2} = \dfrac{{21!}}{{\left( {21 - 2} \right)!2!}} = \dfrac{{21 \times 20}}{2}$
$^{21}{C_2}$= 210 ways
So, the number of possible arrangements of 4 letters is $^4{P_4} = \dfrac{{4!}}{{\left( {4 - 4} \right)!}} = 4 \times 3 \times 2 \times 1 = 24$
So, the total number of ways to select the 4 letters = 24 ways
Thus, the total numbers of words that can be formed is equal to the number of ways selecting 2 vowels $ \times $ number of ways selecting 2 consonants $ \times $ number of arrangements of 4 letters
Therefore, the total number of words of 4 letters = $^{21}{C_2}{ \times ^5}{C_2} \times 4! = 210 \times 10 \times 24 = 50400$
Therefore, the total number of words of 4 letters is 50400 words.
Note: In the above solution we used the method of permutation and combination to get the required result where permutation can be explained as arrangement of an element from the set into a sequence whereas the combination defines the ways of choosing the elements from the set.
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