SOLUTION: The contractor wants to plant six oak trees, nine maple trees, and five poplar trees along the subdivision street. The trees are to be evenly spaced. In how many distinguishable wa

Question 612196: The contractor wants to plant six oak trees, nine maple trees, and five poplar trees along the subdivision street. The trees are to be evenly spaced. In how many distinguishable ways can they be planted?
I got this far:
20!/(6!9!5!) = 77,597,520
I need to know the break down of this equation to get the answer.
Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!

If you have a calculator with a factorial key, then use that.



But if you don't have a factorial key on your calculator, 
then you'll have to do it the following long, long way:

(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
   [(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)][(5)(4)(3)(2)(1)]

Now do some fancy cancelling:

Cancel the middle numbers on the bottom into the last ones on the top

(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
   [(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)][(5)(4)(3)(2)(1)]

Cancel a 5 and a 4 into the 20 on the top:

(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
   [(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)][(5)(4)(3)(2)(1)]

Cancel a 6 and a 3 on the bottom into the 18 on the top:

(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
   [(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)][(5)(4)(3)(2)(1)]

Cancel a 2 on the bottom into the 16 on the top leaving an 8 up there:

                  8
(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
   [(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)][(5)(4)(3)(2)(1)]

Cancel a 5 and a 3 on the bottom into the 15 on the top:

                  8
(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
   [(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)][(5)(4)(3)(2)(1)]

Cancel the 4 on the bottom into the 12 on the top, leaving a 3 up there:

                  8               3
(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
   [(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)][(5)(4)(3)(2)(1)]

Finally cancel the 2 on the bottom into the 14 on the top, leaving a 7 up there:

                  8       7       3         
(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
   [(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)][(5)(4)(3)(2)(1)]

Now the whole denominator has canceled entirely into the numerator,
and we are left with these number to multiply out

(19)(17)(8)(7)(13)(3)(11)(10)

Now get your calculator and multiply that out and get 77597520



Edwin

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