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# How many words can be formed using the letters A thrice, the letter B twice and the letter C once?(a) 60(b) 120(c) 90(d) 59

Last updated date: 25th Jul 2024
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Answer
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Hint: We first try to explain the general formula where we find the arrangement of n things out of which p things are one a kind, q things are other of a kind and the rest are unique as $\dfrac{n!}{p!\times q!}$. For our given problem we find an arrangement of 6 letters out of which 3 A’s are similar, 2 B’s are similar and 1 C is unique. We put the values to find the solution.

Complete step-by-step solution:
We have to construct words using the letters A thrice, the letter B twice and the letter C once.
Therefore, we can use AAABBC.
So, in total there are 6 letters.
Now the formula of arranging n things out of which p things are one a kind, q things are other of a kind and the rest are unique is $\dfrac{n!}{p!\times q!}$.
In our given arrangement we have to find the number of words for 6 letters out of which 3 As are similar, 2 bs are similar and 1 C is unique.
Therefore, the number of words will be $\dfrac{6!}{3!\times 2!}=\dfrac{720}{12}=60$.
The correct option is A.

Note: There are some constraints in the form of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$. The general conditions are $n\ge r\ge 0;n\ne 0$. Also, we need to remember the fact that the notion of choosing r objects out of n objects is exactly equal to the notion of choosing $\left( n-r \right)$ objects out of n objects. The mathematical expression is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}={}^{n}{{C}_{n-r}}$.