Question

In how many ways 7 students can be selected for a math expo from 8 students of ${11^{th}}$ standard, 6 students of ${10^{th}}$ and 4 students of ${9^{th}}$ standard, selecting at least 2 students from each class.

Answer 1

Let's say two students A and B are selected from class ${9^{th}}$. Selecting A and then B or selection B and then A, both are the same. This means the order of selecting students for a group does not matter. Thus, use combinations to find a number of ways of selecting students for the given question.
The number of combinations of $n$ different things taken $r$ at a time, denoted by $C\left( {n,r} \right)$ , is given by
$C\left( {n,r} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
‘AND’ operations are substituted with $' \times '$, implying both combinations are selected together.
‘OR’ operations are substituted with $' + '$, implying either of the one combinations is selected.

Complete step-by-step answer:
Step 1: Given that:
Total number of students selected for a math expo = 7
Number of students of ${11^{th}}$standard = 8
Number of students of ${10^{th}}$standard = 6
Number of students of ${9^{th}}$ standard = 4
Condition: at least 2 students from each class.
Step 2: Types of combinations.
We can select 7 students for the math expo in either of the following ways.
2 students from the class ${11^{th}}$, 2 students from the class ${10^{th}}$, and 3 students from class ${9^{th}}$.
2 students from the class ${11^{th}}$, 3 students from the class ${10^{th}}$, and 2 students from class ${9^{th}}$.
3 students from the class ${11^{th}}$, 2 students from the class ${10^{th}}$, and 2 students from class ${9^{th}}$.
Step 3: Number of combinations.
Number of combinations (or ways) of selecting 7 students $= C\left( {8,2} \right) \times C\left( {6,2} \right) \times C\left( {4,3} \right) + C\left( {8,2} \right) \times C\left( {6,3} \right) \times C\left( {4,2} \right) + C\left( {8,3} \right) \times C\left( {6,2} \right) \times C\left( {4,2} \right)$

$$= \dfrac{{8!}}{{2!\left( 6 \right)!}} \times \dfrac{{6!}}{{2!\left( 4 \right)!}} \times \dfrac{{4!}}{{3!\left( 1 \right)!}} + \dfrac{{8!}}{{2!\left( 6 \right)!}} \times \dfrac{{6!}}{{3!\left( 3 \right)!}} \times \dfrac{{4!}}{{2!\left( 2 \right)!}} + \dfrac{{8!}}{{3!\left( 5 \right)!}} \times \dfrac{{6!}}{{2!\left( 4 \right)!}} \times \dfrac{{4!}}{{2!\left( 2 \right)!}} \\\ = \dfrac{{8 \times 7 \times 6!}}{{2\left( 6 \right)!}} \times \dfrac{{6 \times 5 \times 4!}}{{2\left( 4 \right)!}} \times \dfrac{{4 \times 3!}}{{3!\left( 1 \right)!}} + \dfrac{{8 \times 7 \times 6!}}{{2\left( 6 \right)!}} \times \dfrac{{6 \times 5 \times 4 \times 3!}}{{3 \times 2 \times 1\left( 3 \right)!}} \times \dfrac{{4 \times 3 \times 2!}}{{2\left( 2 \right)!}} + \dfrac{{8 \times 7 \times 6 \times 5!}}{{3 \times 2 \times 1\left( 5 \right)!}} \times \dfrac{{6 \times 5 \times 4!}}{{2\left( 4 \right)!}} \times \dfrac{{4 \times 3 \times 2!}}{{2\left( 2 \right)!}} \\\ = 4 \times 7 \times 3 \times 5 \times 4 + 4 \times 7 \times 5 \times 4 \times 2 \times 3 + 8 \times 7 \times 3 \times 5 \times 2 \times 3 \\\ = 1680 + 3360 + 5040 \\\ = 10080 \\\$$

Final answer: Hence, in 10080 many ways 7 students can be selected for a math expo.

Note: To find the number of handshakes, number of pairs, and selecting team members, always combinations are used.
Another way of calculating the number of arrangements is permutations, denoted by $P\left( {n,r} \right)$ .
The number of permutation of $n$ different objects taken $r$ at a time, where $0 < r \leqslant n$and the objects do not repeat is $n(n - 1)(n - 2)...(n - r + 1)$, which is denoted by $P(n,r).$
$P(n,r) = \dfrac{{n!}}{{(n - r)!}}$
In permutations, the order of selection is important.
$n! = n(n - 1)!$
The notation $n!$ represents the product of first $n$ natural numbers,
i.e. the product $1 \times 2 \times 3 \times ... \times (n - 1) \times n = n!$.
$0! = 1 \\\ 1! = 1 \\\ 2! = 2 \times 1 \\\ 3! = 3 \times 2 \times 1 \\\$