Probability

[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

 

[pka]

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;

I hope that you know binomial notation: \(\dbinom{N}{k}=\dfrac{N!}{(k!)[(N-k)!]}\)
Now the number of ways to place \(N\) identical objects into \(k\) distinct cells is \(\dbinom{N+k-1}{N}\).
In this question are we placing the walls into the colours?

 

[istar]

Thx

 

[istar]

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) = 1,001
For 4 walls: nCr = 14C4 = n!/(n-r)!r! = (14)!/(14-4)!(4)! = (14)!/(10)!(4)!
=(14x13x12x11x10!)/(4x3x2x1)(10)! = (24,024)/(24) = 1,001
So there are there are 1,001 ways Kendra can paint all the walls.

This is how I build it around the equation.

 

[Dr.Peterson]

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

I think the problem is a little ambiguous. Does he have to paint exactly 10 walls the first color, or can it be anything up to 10? Is there a limit on the number of walls painted the second color? Are the two colors both known, or can "another color" be chosen from some list you've omitted?

I'll assume you're right that it will be 10 of color A and 4 of color B.

You are choosing 10 of the 14 walls to have color A.

But once you've done that, you don't have to choose 4 walls again; the 4 that weren't chosen for the first color get the second color. So the answer is 6006; you don't need to calculate it twice.

(However, this does demonstrate an important fact: nCr = nC{n-r}, because each way to choose r elements is also a way to choose the n-r elements that are left out!)

I think pka is interpreting the problem differently, but I'm not sure how.