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An urn contains seven red balls, seven white balls, and seven blue balls. A sample of five balls is drawn at random without replacement. Compute the probability that The sample contains four balls of one colour and one ball of another colour.

Correct answer is $\dfrac{\dbinom31\dbinom74\dbinom21\dbinom71}{\dbinom{21}{5}}$

however I do not understand how it was reached

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    $\begingroup$ The colour represented by four balls can be chosen in $\binom{3}{1}$ ways. For each choice of colour, the actual balls of that colour can be chosen in $\binom{7}{4}$ ways. Continue. $\endgroup$ – André Nicolas Oct 31 '15 at 4:44
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First, pick one of the 3 colors to be the color with four balls -- $3\choose1$

Then from that color pick 4 balls from the 7 -- $7\choose4$

Next pick one of the remaining two colors to be the one with 1 ball -- $2\choose1$

Then pick the 1 ball from that color -- $7\choose1$

Finally, the total number of sets of 5 you could have picked is just 5 from 21 -- $21\choose5$

So you get $$\frac{{3\choose1}{7\choose4}{2\choose1}{7\choose1}}{21\choose5}$$

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