Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
Nov 2011 p62 q3
2865
Geoff wishes to plant 25 flowers in a flower-bed. He can choose from 15 different geraniums, 10 different roses and 8 different lilies. He wants to have at least 11 geraniums and also to have the same number of roses and lilies. Find the number of different selections of flowers he can make.
Solution
Let the number of geraniums be denoted by \(G\), roses by \(R\), and lilies by \(L\). We have the following conditions:
1. \(G + R + L = 25\)
2. \(G \geq 11\)
3. \(R = L\)
From condition 3, we have \(R = L\), so \(G + 2R = 25\).
We consider the possible values for \(G\):
\(G = 11\): Then \(11 + 2R = 25\) gives \(R = 7\) and \(L = 7\). The number of ways to choose the flowers is \(\binom{15}{11} \times \binom{10}{7} \times \binom{8}{7} = 1310400\).
\(G = 13\): Then \(13 + 2R = 25\) gives \(R = 6\) and \(L = 6\). The number of ways to choose the flowers is \(\binom{15}{13} \times \binom{10}{6} \times \binom{8}{6} = 617400\).
\(G = 15\): Then \(15 + 2R = 25\) gives \(R = 5\) and \(L = 5\). The number of ways to choose the flowers is \(\binom{15}{15} \times \binom{10}{5} \times \binom{8}{5} = 14112\).
Adding these possibilities gives the total number of selections:
