omg i dont understand permutations and combinatons?

Question

first of all how does P(4,3)=4*3*2=24? where does the two come from?

and how do you find the value of P(3,3), C(5,5), p(10,4) C(15,6), and P(4,2) these make no sence to me. This stuff is do tommaro and i think im doomed X.X

also im not sure how you can tell something is permutation or combination.

for example: 1. four musical instruments from a group of 12. how do i know which it is.

also how the hell do i do this

6!3!/4!2!

first one too help me actually understand this is who i consider smartest man or girl in the world .

Answer 1

Does it matter in which order you pick your instruments? More than likely NO, so it's a combination and not a permutation. For a permutation, the order matters when selecting.

P(4,3)=4!/(4-3)!=4!/1!
=(4*3*2*1)/1=4*3*2

So suppose three people are lining up and order is important. You can choose your three people from among four people. There are 24 ways to line up 3 people when you can choose from a group of 4 with order being important.

Answer 2

First, definitions; you can find more at the sources.

The quantity of _permutations_ of n objects, k by k, called P(n, k), is the number of ways one can choose k elements of a set of n elements, and the order of these k elements matter.

The quantity of _combinations_ of n elements, k by k, called C(n, k), is the number of ways one can choose k elements of a set of n elements, but the order of the elements does not matter.

Now, formulas. Just use them.

Factorial: This is the "!" operator, always used *after* a number or expression. For n integer > 0, n! = 1 * 2 * ... * n (product of all integers from 1 to n). 0! = 1 and 1! = 1, by convention. Value table, for remembering: 2! = 2, 3! = 6, 4! = 24, 5! = 120.

Permutations: P(n, k) = n! / (n - k)!
Combinations: C(n, k) = n! / ( k! * (n-k)! )

Now, to the specific questions.

> first of all how does P(4,3)=4*3*2=24? where does the two come from?

Formula application: P(4, 3) = 4! / (4-3)! = 24 / 1! = 24/1 = 24. The product 4*3*2 can be seen "laying out" the factorials, and cancelling out factors:

4! / 1! = (4 * 3 * 2 * 1) / 1 = 4 * 3 * 2

> and how do you find the value of P(3,3), C(5,5), p(10,4) C(15,6), and P(4,2) these make no sence to me. This stuff is do tommaro and i think im doomed X.X

I will do only the hardest: C(15, 6). The rest is up to you.

C(15, 6) =
15! / (6! * (15 - 6)! ) =
15! / (6! * 9!) =
(15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (6! * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) =

Cancelling time! I did not expand 6! on purpose, to not confuse things.

(15 * 14 * 13 * 12 * 11 * 10) / 6! =
(15 * 14 * 13 * 12 * 11 * 10) / (6 * 5 * 4 * 3 * 2 * 1) =

Cancelling time, again:

(14 * 13 * 12 * 11 * 10) / (6 * 4 * 2) =
(14 * 13 * 11 * 10) / ( 4 ) =
7 * 13 * 11 * 5 =
5005

> also im not sure how you can tell something is permutation or combination.

> for example: 1. four musical instruments from a group of 12. how do i know which it is.

Rule of thumb: the order of these 4 does matter? If yes, permutation; if no, combination. There are more subtle rules; try asking your teacher.

> also how the hell do i do this

> 6!3!/4!2!

Calculate the factorials (or look up the table) and simplify:

6!3!/4!2! =
(720 * 6) / (24 * 2) =
2160 / 24 =
720 / 8 =
90

Hope this helps.

Answer 3

I will help you with a mneumonic device. In peRmutations, ORDER is important. Now, the word permutations had the letter R as well as oRdeR. So this is your way to associate them together. Now, in combinations, the order is NOT important and the word combination does not have the letter R.

Application to see how to differentiate between the two in solving problems.

1. Combinations. Say you need to pick three members to a student club committee from a group of 10. Note you are just selecting any three.

2. Permutations Say you need to pick three members as President, Vice President and Treasurer. Surely, there is an ORDER. The VP cannot be both VP and President and so on.

Another mneumonic device. For a given set of n, k or n, r
there will always be MORE peRmutations than Combinations. Again the letter R plays prominently MOrE in PeRmutations than in combinations.

Also, In your first example P(4, 3). Here your
n =4 and k =3 and this is calculated as P = n!/(n-k)! =
or, 4!/(4-3)! = 4!/1!

Note that 1! is 1 and 0! is also 1, and 4! = 4x3x2x1 (see where 2 comes from?) compare with 3! = 3x2x1 and 2! = 2x1 and 1! is 1 and nothing else to multiply because 1 is the 'last stop'

This becomes (4x3x2x1) /1 or 24.

Now, Combination C (15,6) means = n!/[(n-k)!(k!)] =

15!/[(15-6)! ][ 6!]
or 15!/(9!)(6!)

Now if you lay it out like in the foregoing paragraph, the portion of 9x8x7.... of the 15! will cancel with the 9! in the denominator. So you are left with

15x14x13x12x11x10/ (6x5x4x3x2x1) = cancel some factors or simplify. Check and see if you have the same answer as 5005


The rest I am sure you can do. Like C(12,4) in your word example.

Challenge yourself. It will give you confidence. Good luck.

Answer 4

lol that sounds truly humorous it would want to correctly be that he change into embarrassed by potential of him being in a washing in structure by potential of the sound of it yet I wouldnt hassle about it too a lot you in uncomplicated words went out 4 2 weeks :p