Find in how many ways an arrangement of $3$ letters can be made from the $26$ different letters of the alphabet if any letter may be used once, twice or thrice. How many of these arrangements will not contain not more than two different letters?
Examples of selections are $\{a,a,a\}$, $\{a,b,b\}$, $\{a,b,c\}$. The first part tells us to count all possible things. The second part says us to count those only $1$st and $2$nd of the examples given above, which means that the chosen set should not have $3$ different letters.
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The total number of arrangements is $26^3$
The number of arrangements with $3$ different letters is $\binom{26}{3}\cdot3!$
The number of arrangements with no more than $2$ different letters is therefore $26^3-\binom{26}{3}\cdot3!$