Question

Find the number of ways of arranging the letters of the word INDEPENDENCE.

Answer 1

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 12 letters out of which 3 As are similar, 2 Bs are similar and 1 C is unique. We put the values to find the solution.

Complete step by step answer:
We have to arrange the letters of the word INDEPENDENCE. So, in total there are 12 letters.
Out of those 12 letters there are 3 N’s, 4 E’s, 2 D’s. Also, there is one each for I, P and C.
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 $\dfrac{n!}{p!\times q!}$.
We apply the same formula for our given problem.
Therefore, the number of arrangements will be $\dfrac{12!}{3!\times 4!\times 2!}=\dfrac{479001600}{288}=1663200$.

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 ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}={}^{n}{{C}_{n-r}}$.