A math question on permutations and combinations?

Question

Why is it that there are 495 ways for 4 people to be chosen out of 12 and also 8 people to be chosen out of 12?

You can try it out on your calculator 12nCr4 and 12nCr8

Thanks!

Answer 1

suppose u have to select 4 persons out of 12

then it can be said in an another way that,

8 of the people are chosen to be removed from 12,and the remaining 4 are chosen



so no. ways of selection of 8------ (to be removed from 12)

=no. of selection of 4----(to be chosen from 12)



and also,vice -versa,



C(12,8)=C(12,4)



generally ways of choosing r objects from n objects,

is equivalent to choosing (n-r) objects from n objects

in the way i said



so C(n,r)=C(n,(n-r))

Answer 2

Consider that the 4 and 8 are opposite sides of the same coin: any time you pick 8 from the 12; you don't pick 4.

For every one of those ways of choosing 8, there is an exactly corresponding way of picking 4.



Also, look at the formula; in the denominator there is an (a-b)! (b!) term [I am probably butchering the nomenclature] so you can see that where you have 12nCr4 and 12nCr8 you have 12!/4!8! and 12!/8!4!, which are equal.



Note that this will not be true for permutations; there are more ways to arrange a larger set than ways to arrange a smaller set. The permutations would be 12!/8! or 12!/4!, which are =/=.





Handy way to remember combinations vs permutations:

Draw a picture of your generic "combination" lock. Write "Combination Lock" underneath it.

Cross out "Combination" and write "Permutation".

In a permutation, order is important: ABC =/= CAB =/=CBA, whereas in a combination order is not important, and ABC = ACB = CAB = etc.

If you scramble the numbers' order, the lock will not open; therefore it is a Permutation lock.

Answer 3

Firstly, this is not a question on permutations AND combinations.

The straight forward answer is to look at the defining equation for the "nCr" function.



n! / [(n-r)! * r!]



The numerator will be the same for both cases, namely 12!.

For the case with 4, the denominator will be (12-4)!4! = 8!4!

For the case with 8, the denominator will be (12-8)!8! = 4!8! = 8!4!



There's a certain symmetry to the situation is another way to describe it. You might try working with smaller numbers and manually picking the combinations. For example, pick n=4, and figure it out for all values of r from 1 to 4.

Answer 4

In the formula n! / r!(n-r)!



if r =4 then n-r = 8, but if r =8 then n-r = 4. So in either case the deminators are the same.



Same goes for all pairs that add up to n

12C3=12C9

12C7=12C7



23C19 = 23C4

Answer 5

12C4

12! / [4!(12 - 4)!]

12! / (4! * 8!)



12C8

12! / [8!(12 - 8)!]

12! / (8! * 4!)



12C0 = 12C12

12C1 = 12C11

12C2 = 12C10

12C3 = 12C9

12C4 = 12C8

12C5 = 12C7

12C6

12C7 = 12C5

12C8 = 12C4

12C9 = 12C3

12C10 = 12C2

12C11 = 12C1

12C12 = 12C0

Answer 6

Try this conceptual explanation. There are twelve people at a party. Four of them are standing in the dining room, and the other eight are in the living room. When you compute the number of possible combinations of four in the dining room, that is equivalent to computing the number of combinations of eight in the living room.