Selecting groups of r items from n items
Example : In how many ways we can select groups of two from 6
students.
Concept :
Let the students be A, B, C, D, E, F . A can group with
each of the remaining five as AB, AC, AD, AE, AF now
we have B, C, D, E, F left, B can group with each of
the remaining four as BC, BD, BE, BF, now we have C,
D, E, F left, C can group with each of the remaining
three as CD, CE, CF, now we have D, E, F left, D can
group with each of the remaining two as DE and DF, E
can group with the last one F , EF. Therefore we have
AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE,
DF and EF, fifteen groups of twos.
Alternatively
One student A can select the other one of his group in
$^5C_1$ ways . If B is another student who does not belong
in the group A, we can select the other one of his group
$^3C_1$ ways the remaining two consists of the third group.
= 5 $\times 3 \times 1$
But as per my understanding :
I used : $^6C_2$ which is equal to 15.
However if I am using this concept in the following question :
Problem :
In how many ways can we divide 12 students in groups
of fours.
A group of 4 can be chosen from 12 students is :
$^4C_{12} = 495$ which is wrong answer and answer is $5775$
Please help on this.. Thanks..
Best Answer
The $12$ students can be lined up in $12!$ ways.
Put in dividers after every fourth student.
For example:
Given $ABCDEFGHIJKL$, we would put in dividers to get $ABCD|EFGH|IJKL$.
We have overcounted, though, in two different ways. Let's examine the resulting problems.
One problem is with regard to the three separate groupings appearing multiple times. In the example above, we would be counting $EFGH|ABCD|IJKL$ a second time. Since there are three groups, we need to divide by $3!$ to avoid this sort of overcounting.
The second problem is that, within each of the three groups, we have four people who are going to appear multiple times. In the example above, we would be counting $CDAB|EFGH|IJKL$ a second time. For each of the groupings, we need to divide by $4!$ to avoid this sort of overcounting. This means dividing by $4!4!4!$ to amelioriate the second issue.
Then our answer becomes:
$$\frac{12!}{3!4!4!4!} = 5,775$$