Possible ways to choose

Question

List all of the possible ways to choose 2 different colors from the list: red, blue, yellow, green. Be systematic. Give some kind of justification.

Answer 1

Hi Lilian N.,

First of all, it's important to know what is meant by "possible ways to choose" when a math problem uses that language. It means, how many different pairs of colors could you choose. A way is just a unique pair of colors (from the list!). Not, ways to choose, in the sense of, "you could flip a coin, or you could ask your friends to pick for you, or you could draw the names of colors out of a hat, or ...."!

So, there is a standard way of attacking this kind of problem (by the way, what you are forming eventually is known as "combinations" of the choices), and it goes like this: you are going to pick the two colors in succession. So, the number of ways to pick the first color is 4, because there are 4 colors you could choose first. But when you have done that, and removed that particular color from the eligible list for the second pick, there are now only 3 colors you could pick second. So far, the total number of ordered picks (permutations) you could have, is 4 x 3 = 12. But, the order you pick the colors in does NOT matter in this problem, because a pair of colors could have been picked in any order. So (yellow, then blue) is the same final pairing as (blue, then yellow), for example. So you divide by 2 (since each pair has 2 possible orders to pick it), to get the final answer, 6.

There is a mathematical shorthand for this problem, and it goes like this. The capital letter "C" represents combinations, and it is followed by a superscript = the total number of items to choose among, and a subscript = the number of items you pick for a "way". So this problem would be: C⁴₂ and it is a particular case of Cⁿ_m == the combinations of n objects taken m at a time. There is a formula for this, and it looks like this: Cⁿ_m = n!

---------------- where the ! symbol refers to taking the factorial of the number directly preceding

(n-m)! m! 4! 4*3*2*1

the symbol. So, in expanded form,C⁴₂ = -------------- = ------------ = 6 , as we just determined.

2! 2! 2*1 * 2*1

Hope this gives you a perspective on the problem. The most difficult part to consider in problems is sometimes, did the questioner mean that the order of the picks matters (permutations), or that it did not matter (combinations)? Or was the questioner clueless (I've seen that happen, too ....)?

--Cheers, -- Mr. d.

Answer 2

Here are the list possibilities: Let's say R=red, B=blue, Y=yellow, G=green.

RB, RY, RG

BY, BG

YG

So there are 6 possibilities!

----

Since we have RB, we can't have BR as another count because there is no order for 2 different colors. Therefore this is COMBINATION because the order doesn't matter and the formula is:

_nC_r= n!/(r! (n-r)!)

₄C₂= 4!/(2!(4-2)!) =4!/(2!*2!) = (4*3*2*1)/(2*1*2*1) = 24/4 = 6.

Answer 3

There are a total of four colors. Assuming that order doesn't matter, we can use the combinations formula,

C(n,r), which means out of n , pick r.

In this case, n = 4 and r = 2.

C(n,r) = n!/((n-r)!r!)

C(4,2) = 4!/((4-2)!2!) = 1×2×3×4/((1×2)(1×2)) = 24/4 = 6.

So there are 6 different ways to choose 2 different colors from red, blue,yellow and green.

Using the first letter of each color:

rb

ry

rg

by

bg

yg

Gives 6 combinations.