Question

The students $${S_1},{\text{ }}{S_2},...{\text{ }}{S_{10}}\;$$ are to be divided into 3 groups A, B, and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities for forming such groups is

Answer 1

Hint: Here the given question is based on concept of combination. To form a 3 groups A, B and C we have to solve it by 3 cases on ways of choosing 1 student to at most 3 students in a group by the concept of combination then summing over all three cases to get a required number of possibilities of forming a group

Formula Used: The formula used to calculate the combination is: $$^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$$----(1)

Complete step by step Solution: Given, the 10 students $${S_1},{\text{ }}{S_2},...{\text{ }}{S_{10}}\;$$ i.e., $$n = 10$$
To form a 3 groups A, B and C group has at least one student and the group C has at most 3 students. Whenever,
Case 1: If group C has one student
Number of ways of selecting one student who will be in group C is $${}^{10}{C_1}$$ and the remaining students will be distributed into 2 groups is $$\left( {{2^9} - 2} \right)$$
$$ \Rightarrow \,{}^{10}{C_1}\left( {{2^9} - 2} \right)$$
Case 2: Similarly, if group C has two students
$$ \Rightarrow \,{}^{10}{C_2}\left( {{2^8} - 2} \right)$$
Case 3: Similarly, if group C has three students.
$$ \Rightarrow \,{}^{10}{C_3}\left( {{2^7} - 2} \right)$$
Now, Total number of ways is
$$ \Rightarrow \,{}^{10}{C_1}\left( {{2^9} - 2} \right) + {}^{10}{C_2}\left( {{2^8} - 2} \right) + {}^{10}{C_3}\left( {{2^7} - 2} \right)$$
By the formula of combination
$$ \Rightarrow \,\dfrac{{10!}}{{\left( {10 - 1} \right)!1!}}\left( {512 - 2} \right) + \dfrac{{10!}}{{\left( {10 - 2} \right)!2!}}\left( {256 - 2} \right) + \dfrac{{10!}}{{\left( {10 - 3} \right)!3!}}\left( {128 - 2} \right)$$
$$ \Rightarrow \,\dfrac{{10!}}{{9! \cdot 1!}}\left( {510} \right) + \dfrac{{10!}}{{8! \cdot 2!}}\left( {254} \right) + \dfrac{{10!}}{{7! \cdot 3!}}\left( {126} \right)$$
$$ \Rightarrow \,\dfrac{{10 \times 9!}}{{9!\, \times 1!}}\left( {510} \right) + \dfrac{{10 \times 9 \times 8!}}{{8!\, \times 2!}}\left( {254} \right) + \dfrac{{10 \times 9 \times 8 \times 7!}}{{7!\, \times 3 \times 2!}}\left( {126} \right)$$
$$ \Rightarrow \,10\left( {510} \right) + 45\left( {254} \right) + 120\left( {126} \right)$$
$$ \Rightarrow \,5100 + 11430 + 15120$$
$$ \Rightarrow \,31650$$

Hence, $$31650$$ ways of possibilities of forming such groups are there

Note: Remember, factorial is the continued product of first n natural numbers is called the “n factorial” and it represented by $n! = \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right).....3 \cdot 2 \cdot 1$.
the student has to know the difference between at least and at most. The word “at least” means it will be the minimum value and then it can be exceeded to the maximum. The word “at most” means it is the maximum value and it can’t exceed further, we have to consider the minimum value also.