I had a question which is as follows:Number of words of 4 letters formed using the word IITJEE.The book says the answer as coefficient of $x^4$ in 4!$\mathrm{[1+ \frac {x}{1!}+\frac{x^2}{2!}]}^{2}[1+x]^2$.My question is where did this $\mathrm{[1+ \frac {x}{1!}+\frac{x^2}{2!}]}^{2}$ come from?
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see exponential generating functions. – abstractnature Feb 15 '14 at 06:09
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$\mathrm{[1+ \frac {x}{1!}+\frac{x^2}{2!}]}^{2}$ means the generating function for 2 I's and 2 E's here, as they are alike. – abstractnature Feb 15 '14 at 06:19
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The individual factors account for the number of each distinct letter: $$4!\overbrace{\left(1+\frac{x}{1!}+\frac{x^2}{2!}\right)}^{2 \times \text{I}}\overbrace{\left(1+\frac{x}{1!}+\frac{x^2}{2!}\right)}^{2 \times \text{E}}\overbrace{\left(1+\frac{x}{1!}\right)}^{\text{T}}\overbrace{\left(1+\frac{x}{1!}\right)}^{\text{J}}$$
For example, if we have $1$ letter I in the $4$-letter word, it's accounted for by the $x$ term in the first factor: this uses up one of the $x$'s in $x^4$ (informally). If we instead have $2$ letter Is in the $4$-letter word, then this is instead counted by $x^2$ term in the first factor, using up two of the $x$'s in $x^4$.
Given four letters, there's $4!$ (not necessarily distinct) ways of ordering them, and we divide by $k!$ for each letter we have $k$ copies of to account for non-distinct orderings (whence the coefficients). This essentially recovers the multinomial coefficient.
Rebecca J. Stones
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We divide by k! because we are using exponential generating function rather than linear generating function, because it will account for permutations, as linear accounts for combinations. – abstractnature Feb 15 '14 at 06:24