A florist has to make a floral arrangement. She has 6 Banksias, 5 wattles and 4 Waratahs. All the flowers of each kind are different. In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind? Please show working.
I have tried to use the combinations formula but have gotten two different answers - 1200 and 4800 depending on whether I do the first 9 and then the last flower or all at once.
Just an interesting problem I know and have not been able to figure out.
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$\begingroup$At least three of each kind means that $10=4+3+3=3+4+3=3+3+4$ and therefore the number of different bunches of $10$ flowers out of $6$ Banksias, $5$ Wattles and $4$ Waratahs should be$$\binom{6}{4}\binom{5}{3}\binom{4}{3}+\binom{6}{3}\binom{5}{4}\binom{4}{3}+\binom{6}{3}\binom{5}{3}\binom{4}{4}=600+400+200=1200.$$We get $4800$ if we choose the first $9$ and then we add the last one (the fourth of the same kind): $$\binom{6}{3}\binom{5}{3}\binom{4}{3}\cdot (3+2+1)=4800$$But here we overcount each bunch four times because each of the four flowers of the same kind can be the last one. So we have to divide $4800$ by $\mathbf{4}$ and we obtain again the correct number $1200$.
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