Permutations and Combinations are CRAZY HARD. (1 Viewer)

[morganforrest]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

This is just a simple probability question. How many ways can jan be selected (4/X) times the number of ways her mother can (3/Y)

Where X is the sample space, the number of students applying for the committee in total
and Y is the number of parents applying

 

[webby234]

Total possible committees:
12C5 * 25C4 * 7C3

Committees that involve a specific student and her parent:
Of the remaining 6 parents, 2 of them get chosen. Of the remaining 24 students, 3 of them get chosen, and 5 of the 12 teachers are still to be selected.

So it's 12C5 * 24C3 * 6C2

= 12/175

Alternatively, the chances of getting selected for the parent is 3/7 and for the student is 4/25. Total is 3/7 * 4/25 = 12/175

 

[m.incognito]

5 men 5 women are to be arranged in a circle so that the men are separated.
In how many ways can this be done if 2 particular women must not be next to a particular man?

so... would you take the complementary of them being together?

help!

 

[kubi]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

omg~~ im doing it at school at the moment...
im so lost...T_T
it's like the hardest topic...along with parametrics...=="

 

[m.incognito]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

^ ahem question above. and.

The ratio of the number f combinations of (2n+2) different objects taken n at a time to the number of combinations of (2n - 2) different objects taken n at a time is 99:7 Find the value of n.


HOW THE EFF DO I DO THAT?!

 

[iEdd]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

The ratio of the number f combinations of (2n+2) different objects taken n at a time to the number of combinations of (2n - 2) different objects taken n at a time is 99:7 Find the value of n.

Maybe:
Let ⁽²ⁿ⁺²⁾C_n = 99 and ^(2n-2)C_n = 7
Remembering ⁿC_r = ^n!/_(n-r)!r!

 

[Kujah]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

The ratio of the number f combinations of (2n+2) different objects taken n at a time to the number of combinations of (2n - 2) different objects taken n at a time is 99:7 Find the value of n.


HOW THE EFF DO I DO THAT?!

Ah, Fitzpatrick right?

So we have
[(2n +2) C (n)] / [(2n - 2) C (n)] = 99/7

Expand the combination notation

[(2n + 2)!/(2n + 2 - n)! n!] / [(2n - 2)! / (2n - 2 - n)! n!] = 99/7

To start simplifying on the right side, I will flip the second fraction and multiply.

[(2n + 2)! / (n +2)! n!] x [(n - 2)! n! / (2n - 2)!] = 99/7

Expand the factorials and you'll be able to cancel.

[(2n + 2)(2n +1)(2n)(2n - 1)(2n - 2)! / (n+2)(n + 1)(n)(n - 1)(n - 2)! n!] x [(n - 2)! n! / (2n -2)!] = 99/7

The ending factorials from the first fraction cancel out with the factorials on the second fraction and we are left with.

2(n +1)(2n + 1)(2n)(2n - 1) / (n + 2)(n + 1)(n)(n -1) = 99/7

Reduce further by cancelling terms in the numerator with those in the denominator.

4(2n +1)(2n - 1) / (n + 2)(n - 1) = 99/7

(16n^2 - 4) / (n^2 + n - 2) = 99/7

Cross multiply to get rid of fractions

112n^2 - 28 = 99n^2 + 99n - 198

Get everything on one side set equal to zero.

13n^2 - 99n + 170 = 0

Use the quadratic formula to solve for n.

n = [99 +or- sqrt( 99^2 - 4(13)(170)] / 2(13)

n = [99 +or- sqrt(961)]/26

n = 5 or n = 34/13

Therefore, n=5 'cause its an integer.

 

[iEdd]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

Geez, do it all for her why don't you.

 

[SoulSearcher]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

Geez, do it all for her why don't you.

I remember a time when I used to be that kind of person.[comment]Whether I still am or not depends on my mood at the time.[/comment]

 

[m.incognito]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

I think I just imploded... Yes kevin, Fitzpatrick

I hate you, maths.