Formation of Groups - Practice Questions & MCQ

Consider that 12 people have to be divided among three groups of unequal sizes such as one group has 3 members, one group has 4 members and one group has 5 members.

We could have formed a group of 3 members in ${ }^{12} \mathrm{C}_3$ ways. Having formed a group of three, we would be left with $12-3=9$ people. A group of 4 members can be formed from these 9 members in ${ }^9 \mathrm{C}_4$ ways. For each group of 3 members formed earlier, there would be further ${ }^9 \mathrm{C}_4$ ways of forming a group of four. Thus, the total possible number of ways of forming a group of 3 and a group of 4 would be ${ }^{12} \mathrm{C}_3 \times{ }^9 \mathrm{C}_4$. Now there would be 5 people left who are the third group i.e. the third group can be formed in only 1 way. To maintain consistency, we will say that the third group can be formed in ${ }^5 \mathrm{C}_5$ ways (which is 1 anyway). Thus, the total number of ways of forming the groups is ${ }^{12} \mathrm{C}_3 \times{ }^9 \mathrm{C}_4 \times{ }^5 \mathrm{C}_5$. On expansion, this equals $\frac{(12)!}{3!4!5!}$.

This concept can be generalized for $(m+n+r)$ distinct objects which have to be grouped into three unequal groups containing $m, n$, and $r$ objects. So this grouping can be done in $\frac{(\mathrm{m}+\mathrm{n}+\mathrm{r})!}{\mathrm{m}!\mathrm{n}!\mathrm{r}!}$ number of ways

This same concept will apply for $(m+n)$ distinct object which has to be grouped in two unequal containing $m$, and n items.

Number of ways of dividing mn object into m groups such that all groups contain n objects

equals $\frac{(\mathrm{mn})!}{(\mathrm{n}!)^{\mathrm{m}}} \times \frac{1}{\mathrm{~m}!}$

Example: How many ways 12 people can be divided into 3 groups, such that all three of them contain 4 people each?

Solution: The number of ways of forming the three groups is ${ }^{12} \mathrm{C}_4 \cdot{ }^8 \mathrm{C}_4 \cdot{ }^4 \mathrm{C}_4$. Since we are multiplying these three factors, we are inadvertently also arranging the groups in a particular order. (Remember that if one position can be filled in 5 ways, another can be filled in 4 ways, and the third can be filled in 3 ways when we apply the rule of AND i.e. 5 × 4 × 3, we are basically finding the number of “arrangements” of the three positions) 

But the question requires us to just form groups and we do not have to “arrange” the groups. Since we have arranged 3 objects which did not have to be arranged, we have counted each unique way of forming the groups 3! times i.e. 6 times. Thus, the correct answer would be found by dividing the earlier found answer by 3! This will give you the above formula itself.