How many $5$ letter arrangements can be made using the letters of the word INDEPENDENT?

Question

score 3 · Answer 1 · answered Oct 08 '15 at 10:09

3

The word has $3$ each of $N$ and $E$, $2$ of $D$, $1$ each of $I, P. T$

There are two ways to do it. The first is using a formula:

coefficient of $x^5$ in $5!(1+x)^3(1+x+x^2/2!)(1+x+x^2/2! +x^3/3!)^2 = 3320$

You should be able to see how it takes care of repetitions of letters.

The second way is a long one, classifying by types:

$3-2$ of a kind: $\binom21\binom21\frac{5!}{3!2!} = 40$

$3-1-1$ of a kind: $\binom21\binom52\frac{5!}{3!} = 170$, and so on upto

$1-1-1-1-1$ of a kind: $\binom655! = 720$

answered Oct 08 '15 at 10:09

true blue anil

Does Generetaing function method takes care of arrangements? can you explain why you multiplied with $ 5!$ – Ekaveera Gouribhatla Jun 15 '18 at 03:11
@Ekaveera Kumar Sharma. See my answer here https://math.stackexchange.com/questions/1587851/number-of-ways-of-forming-4-letter-words-using-the-letters-of-the-word-ramana/1588047#1588047 – true blue anil Jun 15 '18 at 06:17

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