Question

Mary has $6$ identical basil plants, and three different window sills she can put them on. How many ways are there for Mary to put the plants on the window sills

Answer 1

There are 28 ways for Mary to put the plants on the sills.

We can use the Binomial Theorem to solve this problem.

We have 6 plants and 3 sills, so we have6 + 3 − 1 = 8 total objects.

Since we have 8 objects, we can use the Binomial Theorem to expand $(x + y)^{8}$.

The coefficient   $x^{6} y^{2}$ will be the number of ways for Mary to put the plants on the sills. We can expand $(x + y)^{8}$

using the Binomial Theorem:

$(x + y)^{8}=(\left\ {{8}\atop {0}} \right. )x^{8}+ (\left\ {{8}\atop {1}} \right. )x^{7}y+(\left\ {{8}\atop {2}} \right. )x^{6}y^{2} +(\left\ {{8}\atop {3}} \right. )x^{5}y^{3} +(\left\ {{8}\atop {4}} \right. )x^{4}y^{4}+(\left\ {{8}\atop {5}} \right. )x^{3}y^{5} +(\left\ {{8}\atop {6}} \right. )x^{2}y^{6} +(\left\ {{8}\atop {7}} \right. )x^{1}y^{7} +(\left\ {{8}\atop {8}} \right. )y^{8}$

Since we are only interested in the coefficient of $x^{6} y^{2}$ , we can ignore all terms that do not have $x^{6}$ and  $y^{2}$.

Therefore, we are left with $(\left\ {{8}\atop {2}} \right. )x^{6}y^{2}$

$(\left\ {{8}\atop {2}} \right. )=\frac{8!}{2!(8-2)!} =\frac{8!}{2!6!} =\frac{8*7*6!}{2!6!} =28$

Therefore, there are 28 ways for Mary to put the plants on the sills.

Learn more about Binomial theorem here brainly.com/question/2584994

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