7 orchids can be lined as 7!. This means that for the first orchid of the line, you can select 7 options. When you place the first orchid, for the second option you can select among 6 since 1 orchid has already been placed. Similarly, for the 3rd orchid of the line, you have left 5 options. The sequence goes in this fashion and for 7 orchids, you have 7*6*5*4*3*2*1 possibilities. However, there is a restriction here. 3 of the orchids are white and 4 are levender. This means that it does not make a difference if we line 3 white orchids in an arbitrary order since it will seem the same from the outside. As a result, the options for lining the 7 orchids diminish. The reduction should eliminate the number of different lining within the same colors. Similar to 7! explanation above, 3 white orchids can be lined as 3! and 4 levender orchids can be lined as 4!. To eliminate these options, we divide all options by the restrictions. The result is:

$\frac{7!}{4!*3!}$ = 35. [(7*6*5*4*3*2*1/(4*3*2*1*3*2*1)]

3 0

3 years ago

What's the full answer

Jet001 [13]

????????????????????????

8 0

3 years ago