# Regarding Combination Question

• Jun 9th 2013, 09:18 AM
Regarding Combination Question
Good Evening,

Hey guys I have a question here:

Two drugs are to be tested on 60 lab mice. Each mouse receives either one drug or acts as a control. Drug A is to be tested on 22 of the mice. Drug B is to be tested on another 22 of the mice. The remaining 16 mice are to act as a control group.How many different ways can the tests performed?

the answer provided by my teacher is :

60!/( 22! 22! 16! )

but may i know why?....can someone explain it?

Thanks very much
• Jun 9th 2013, 09:31 AM
Plato
Re: Regarding Combination Question
Quote:

Two drugs are to be tested on 60 lab mice. Each mouse receives either one drug or acts as a control. Drug A is to be tested on 22 of the mice. Drug B is to be tested on another 22 of the mice. The remaining 16 mice are to act as a control group. How many different ways can the tests performed?
the answer provided by my teacher is :
60!/( 22! 22! 16! )
but may i know why?....can someone explain it?

If you have the string $\displaystyle AAAAABBBBBNNN$ there are $\displaystyle \frac{13!}{(5!)^2(3!)}$ ways to rearrange that string.

In your case there are 22A's, 22B's & 16N's.
• Jun 11th 2013, 12:13 PM
HallsofIvy
Re: Regarding Combination Question
It is hard to know how to justify that answer without knowing what you DO know about combinatorics. Can we assume that you know that n! is defined as n(n-1)(n-2)...(3)(2)(1)?

First, do you know that there are 60! ways of arranging 60 different things? There are 60 ways of deciding which to put first, then 59 left so there are 59 ways to choose the second. There are 60(59) ways to choose the first two. That leaves 58 different objects so 58 ways of choosing the third, 57 ways to choose the fourth, etc. When you get down to only two objects left, there are, of course, 2 ways to choose the next to last object and then just 1 way to choose the last. That is, there are 60(59)(58)(57)...(3)(2)(1)= 60! ways to arrange 60 different things.

But not all 60 "objects" are different here. Writing "A" for drug A, "B" for drug B, and "N" for neither, as Plato suggests, there are 22 "A"s, 22 "B"s, and 60- 22- 22= 16 "N"s.
If those were all different, if, for example, we labeled the "A"s as "A1", "A2", etc, and the same for the "B"s and "N"s, there would be 60! ways of arranging them. But there are 22! ways of rearranging just the "A"s so we don't want to count strings in which only the different "A"s are swapped as different. We need to divide by 22! to discount that. The same thing happens for the 22 "B"s and 16 "N"s so we need to divide by 22! again and by 16!. That is where the $\displaystyle \frac{60!}{22! 22! 14!}$ comes from.

(As a simple example, if there were 5 letters, two "A"s, two "B"s, and one "N", if we label them $\displaystyle A_1A_2B_1B_2N$ there would be 5 distinct "objects" and so 5!= 5(4)(3)(2)(1)= 120 ways of arranging them. But, in fact, we don't want to treat, say, $\displaystyle A_1A_2B_1B_2N$ as different from $\displaystyle A_2A_1B_1B_2N$. Since, for any given arrangement of the "B"s and "N" we don't want to count the 2!= 2 ways of rearranging only the "A"s as different, we divide by that. Similarly for the two "B"s. We don't really need to "divide by 1" to allow for the single "N" but for completeness we can write this as $\displaystyle \frac{5!}{2!2!1!}= \frac{120}{4}= 30$. There are 30 different ways to arrange "AABBN".