Question

If there are 60 students that shall work in groups of 2 and 3 students and you want to have 12 groups of each, how many variations are there to put the 24 groups together?

I'm a little bit confused, but I would go for something like n!/(n-k!) but then I don't know what are my k's. If these are the groups, how can I bring in the scale of the two different group types?

Answer 1

\[_{60}C_{24}\cdot _{24}C_2 \cdot _{36}C_3\]

Answer 2

first is number of way of splitting 60 students into the 2 groups (working in pairs or working in groups of 3)

2nd number is number of possible groups of 2 from 24 students.

3rd number is number of groups of 3 from 36 students.


does that make sense to you?

Answer 3

So for \[_{60}C _{24}\] my calculation would be \[\frac{ 60! }{ (60-24)! }\] ?

Answer 4

\[_nC_r = \frac{ n! }{ (n-r)!\,\cdot r! }\]

Answer 5

or\[_nC_r = \frac{ _nP_r }{ r! }\]

Answer 6

With P as the probability?

Answer 7

no P is permutation...\[_nP_r=\frac{ n! }{ (n-r)! }\]

Answer 8

But that calculation would only fit for small numbers won't it? If I raise the number of students for example to 120, my calculator says: D'oh! :D

Answer 9

yes, the numbers do get big quite quickly!

Answer 10

So that I can't calculate it for 120? Because task b) says "How many posibilitys are there, if you double the amount of students?"

Answer 11

@pgpilot326 Shouldn't it be
\[64C24*24C2*36C3+60C24*24C3*36C2?\]

Answer 12

you can still express it using combinations even though you won't get a number.

and @M0j0jojo I don't think so...

if you take 24 and choose 3, you only get 8 groups and for 36 choose 2 you get 18 groups.

Answer 13

oh wait, im not thinking right

Answer 14

So how would a combination for that look like?

Answer 15

@ganeshie8

Answer 16

you can just use the shorthand...
\[\left( _{60}C_{24}\right)\cdot\left( _{24}C_2 \right)\cdot\left( _{36}C_3\right)\]if you double the number of students to 120, all of the numbers double except the 2 and the 3 (the number of students in the groups remains the same, i.e. groups of 2 and groups of 3).

Answer 17

you can also write combinations like this\[_nC_r=\left(\begin{matrix}n \\ r\end{matrix}\right)\]

Answer 18

Ah ok. Thank you! I'm always a little bit sad, if there is no number at the end of my calculations :(

Answer 19

\[\left(\begin{matrix}60 \\ 24\end{matrix}\right)\cdot \left(\begin{matrix}24 \\ 2\end{matrix}\right)\cdot \left(\begin{matrix}36 \\ 3\end{matrix}\right)\]

Answer 20

you could try an online calculator...

Answer 21

\[\left(\begin{matrix}120 \\ 48\end{matrix}\right)=8800131833100474735688053494121225\]

Answer 22

that's according to the calculator on this site:
http://www.calculatorsoup.com/calculators/discretemathematics/combinations.php

Answer 23

Thank you very much!

Answer 24

you're welcome!