Question: Find the number of ways of forming 4 letter words using the letters of the word "RAMANA"

This can be solved easily by taking different cases.

  1. All 3 'A's taken: remaining one letter can be chosen in $^3C_1$ ways. Total possibilities $=^3C_1\cdot\frac{4!}{3!}=12$
  2. Only 2 'A's taken: remaining two letters out of {R,M,N} can be chosen in $^3C_2$ ways. Total possibilities $= ^3C_2\cdot\frac{4!}{2!}=36$
  3. Only one A: Number of ways: $4!=24$

Total $=72$.

But my teacher solved it like this. He found the coefficient of $x^4$ in $4!\cdot(1+\frac{x}{1!})^3(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!})$ which also came out to be $72$.

Why does this work? Also, if I avoid the factorials, I get number of combinations. That is, number of combinations $=$ coefficient of $x^4$ in $(1+x)^3(1+x+x^2+x^3)$

  • $\begingroup$ The equation which you have written is of multinomial theorem so you got it right $\endgroup$ Dec 24, 2015 at 16:30
  • $\begingroup$ Can you suggest a book or website where I can learn multinomial theorem? $\endgroup$
    – Aditya Dev
    Dec 24, 2015 at 16:40
  • $\begingroup$ Check it out on wikipedia or wolframalpha $\endgroup$ Dec 24, 2015 at 16:43
  • $\begingroup$ My question was not to check if its correct, but to know why it works or The logic behind it. $\endgroup$
    – Aditya Dev
    Dec 24, 2015 at 16:44
  • 1
    $\begingroup$ Your professor is using exponential generating functions. $\endgroup$ Dec 24, 2015 at 18:40

1 Answer 1


Suppose you have a $4$ letter string composed of, say, $1$ distinct and $3$ identical letters

There would be $\frac{4!}{1!3!}$ permutations, also expressible as a multinomial coefficient, $\binom{4}{1,3}$

Similarly, for $2$ distinct, $2$ identical, and $3$ distinct, $1$ identical,
it would be $\binom{4}{2,2}\;$ and $\binom{4}{3,1}$ respectively.

In the polynomial expression $4!(1+x/1!)^3(1+x+x^2/2!+x^3/3!)$,
the 4! corresponds to the numerator, whatever the combination; the first term in $( )$ corresponds to choosing one or more from $R,M,N$; and the other term corresponds to choosing $1,2,$ or $3 A's$

It will become evident why this approach works if we expand the first term in ( ), and compare serially with your case approach by just using the appropriate coefficients to get terms in $x^4$

$4!(1 + 3x + 3x^2 + x^3)(1 + x + x^2/2! + x^3/3!)$

To find the coefficient of $x^4$, consider the three cases that produce $x^4$

One from $R,M,N, 3A's : 4!\cdot3\cdot\frac{1}{3!} = 12$
Two from $R,M,N, 2A's : 4!\cdot3\cdot\frac1{2!} = 36$
Three from $R,M,N, 1A : 4!\cdot1\cdot 1 = 24$

Coefficient of $x^4 = 12+36+24 = 72$

We can now clearly see why the coefficient of $x^4$ in the expression automatically gives all possible permutations of $4$ letters


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