We could also use generating functions (which has recently been my favorite method since I suck at identifying cases).

You are watching: How many permutations of 4 letters

Consider the multiset $M=<1 extE, 1 extX, 2 extA, 1 extM, 2 extI, 2 extN, 1 extT, 1 extO>$.

The number of $4$-permutations of $M$ is the coefficient of $x^4/4!$ in the exponential expansion of

$$(1+x)^5 left(1+x+dfracx^22! ight)^3.$$

The coefficient of $x^4$ above is $409/4$. Multiplying by $4!$ we get $2454$.

The $(1+x)$ represent the letters that are not repeated, while $(1+x+x^2/2)$ represents those letters that are repeated twice.


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The word examination has $11$ letters. It is composed of $8$ different letters, $3$ of which repeat once.

As Andre suggests, break it down into three cases.

All different. Choose the four letters out of the eight you"ll use $8 choose 4$, and then order them ($4!$).One double. Choose which of the three letters is the double $3 choose 1$, then choose which positions they go in the word $4 choose 2$. Then, choose the two remaining letters $7 choose 2$, and decide which goes first and which goes second in the remaining two spaces ($2!$).Two doubles. Choose which two of the three doubled letters to use $3 choose 2$, then order them ($6$).

Adding all of these gives $2454$.


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