Combinations problem

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Loki123

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Sep 22, 2021
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How do I get n from this? IMG_20220404_002158.jpg
When I write it out I get 4=(n-3)....(n-11).
Solving that would take me yeaaars.
 
Biggest Hint so far: What does k and n-k add up to? .......................................edited
 
Last edited by a moderator:
Steven G said:
Another hint: If you choose 12, then you are not choosing 3.
Click to expand...
Okay so I do get it but I also have a bazillion questions.

n-3=12
n=15 (correct)
but why not:
3=n-3
n=6
or
12=n-12
n=24
or (although impossible)
n-3=n-12
0=-9
 
Loki123 said:
3=n-3
n=6
Click to expand...
Hi Loki,
I think you're mixing up your unknowns k vs. (n-k). There are 2 ways you could've define k.
Case 1: k=3
For k=3, then n-k=n-3=12 -> n=15

Case 2: k=12
For k=12, then n-k=n-12=3 -> n=15

Either way it works. Hope this clear up your confusion :)
 
Loki123 said:
Okay so I do get it but I also have a bazillion questions.

n-3=12
n=15 (correct)
but why not:
3=n-3
n=6
or
12=n-12
n=24
or (although impossible)
n-3=n-12
0=-9
Click to expand...
The point is that the number of ways to choose three from fifteen is the same number as choosing twelve from fifteen.
[imath]\dbinom{15}{3}=\dfrac{15!}{(3!)(15-3)!}=\dfrac{15!}{([15-12]!)(12)!}=\dbinom{15}{12}[/imath]

[imath][/imath][imath][/imath]
 
The idea is based on the symmetry of Pascal's Triangle:

1649097228348.png

The green cells are [imath]\dbinom{15}{3}[/imath] and [imath]\dbinom{15}{12}[/imath].

You wanted to solve [imath]\dbinom{n}{3}=\dbinom{n}{12}[/imath], which are on the nth row of the triangle. The only two cells in a row that are equal are symmetrical, so you are looking for a row in which entries 3 and 12 are the same distance from the ends. That is, n-3 = 12.
 
Suppose you were going to hire 99 people from 100 applicants.

These 100 people are in a large office and you have to pick the 99 from the 100.

You can do this in two ways.

1) You can pick out 99 people one by one and ask them to go to the other corner of the room. Of course, after doing so there will be 1 person left. The 99 people in the other corner of the room are the ones you chose to hire.

2) You can pick 1 person Not to hire and ask this person to go to the other corner of the room. This time 99 people will be left--the 99 you want to hire.

This is saying that 100C99 = 100C1 or in general nCr = nC(n-r). This is saying that the number of ways to choose r items from n items is exactly the same number of ways of (not) choosing n-r items from n items.
 
Or: mathematically,

\(\displaystyle \dbinom{n}{r} \ = \ \frac{n!}{(r!) * (n - r)!} \) ............. AND

\(\displaystyle \dbinom{n}{n-r} \ = \ \frac{n!}{(n-r)! * [n - (n-r)]!} \ \ = \ \frac{n!}{(n-r)! * r!}\)
 
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