In how many ways can Ben paint his apartment?

[Istar]

The rooms of Ben’s apartment has 14 walls. He has enough paint to cover 10 of these walls with one color and the rest with another color. In how many ways could Ben paint his apartment ?

This problem I used combinations for 10 walls for one color and 4 walls for one color; Here my calculations:

14 walls in total, 10 walls in one colour and the rest (4 walls) in another colour.

For 10 walls: nCr = 14C10 = n!/(n-r)!r! = (14)!/(14-10)!(10)! = (14)!/(4)!(10)!

=(14x13x12x11x10!)/(4x3x2x1)(10)! = (24,024)/(24) = 6,006

For 4 walls: nCr = 14C4 = n!/(n-r)!r! = (14)!/(14-4)!(4)! = (14)!/(10)!(4)!

=(14x13x12x11x10!)/(4x3x2x1)(10)! = (24,024)/(24) = 6,006

i wonder if I am in the right direction ? Pls and Thank you

 

[MarkFL]

I think you are definitely on the right track. We could either look at the number of ways to choose 10 from 14 or the number of ways to choose 4 from 14, as after all these will necessarily return the same number since:

$${n \choose r}={n \choose n-r}$$

$$\frac{n!}{r!(n-r)!}=\frac{n!}{(n-r)!(n-(n-r))!}$$

$$\frac{n!}{r!(n-r)!}=\frac{n!}{r!(n-r)!}\quad\checkmark$$

$${14 \choose 4}=\frac{14!}{4!(14-4)!}=\frac{14!}{4!10!}=\frac{14\cdot13\cdot12\cdot11}{4!}=7\cdot13\cdot11=1001$$

Your work was correct up until the final division where you made a slight error.

 

[Istar]

Thank you, guess did most of the work and didnt finish it . So do I have to multiply 14C10 x 14C4 to get the final answer ?

 

[MarkFL]

No, the final answer is that there are 1,001 ways for Ben to paint his apartment. He can either choose the 10 walls for one color, or the 4 walls for the other. In either case he will find there are 1001 ways to do so, and that they are equivalent.

 

[Istar]

Thank you very much