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From 3 capitals, 5 consonants, and 4 vowels, how many words can be made, each containing 3 consonants and 2 vowels, and beginning with a capital?

Last updated date: 19th Jun 2024
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Hint: In this question, we have to make a 6 letter word. So, we have six places to fill. First, we will choose one capital letter from 3 capital letters, then choose 3 consonants from 5. And then choose 2 vowels from 4. Finally, the letters at these six places to get the answer.

Complete step-by-step solution:
We have in question 3 capital letters, 5 consonants, and 4 vowels.
First, we will find ways of choosing the numbers. Here we use the combination formula for choosing the numbers as: ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $ and for arranging n elements at n places, we use $n!$.
The number of ways of choosing 3 consonants from 5 consonants =${}^5{C_3} =10$ ways.
The number of ways of choosing 2 vowels from 4 vowels =${}^4{C_2} = 4$ ways.
The number of ways of choosing 1 capital from 3 capitals =${}^3{C_1} = 3$ ways.
We have to form a six-letter word where the first letter is capital and the remaining five can be anything out of vowels and consonants.
So, the first position is fixed by capitals and in the remaining 5 places the consonants and vowels can be arranged in $5!$ ways.
We know that $5! =120.$
$\therefore $ Total number of words that can be formed = $10 \times 6 \times 3 \times 120 = 21600$ ways.

Note: In this type of question, we use two steps to solve the question. In the first step, we use combinations to choose the letters and in the second step, we arrange the letters using the permutation formula. The number of combination of ‘n’ different things taken ‘r’ at a time is given by:
${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} = \dfrac{{{}^n{P_r}}}{{r!}}$.