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.

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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