We interpret the question as saying we cannot have two (or three) O's together. Think of the slots occupied for the remaining $5$ letters. There are $6$ spaces "between" these slots for the O's to be squeezed into, no more than one O per space. Here the number of spaces is $6$ because I am counting the two endspaces. We choose $3$ of these $6$ spaces for the O's. This can be done in $\dbinom{6}{3}$ ways. For each of these ways, the T can be placed in $5$ ways, then the M in $4$ ways, then the W in $3$ ways. Now it is all done, the R's take the remaining two slots. So our count is $$\binom{6}{3}(5)(4)(3).$$ Remark: There are many other ways of counting. The advantage of this one is that it generalizes smoothly to a situation where the length of the word, and the number of O's, is much larger. The idea can be adapted for similar problems. A standard one is to ask how many ways can we line up $9$ adults and $5$ children in a row if no two childen can be next to each other. |
In how many ways can the letters of the word psychology be arranged so that the two Os
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