Combinations

Combinations refer to a selection of objects in which the order is not important.

Combinations of \(n\) distinct objects taken \(r\) at a time can be represented algebraically as such:

\(_nC_r = \cfrac{_nP_r}{r!}\)

\(= \cfrac{n!}{(n-r)! \cdot r!}\)

\(= \cfrac{n!}{(n-r)!r!}\)

The symbols \(C_n^r\), \(C(n, r)\), and \(\binom{n}{r}\) all represent the same mathematical concept, known as combinations, which is used to determine how many different ways you can choose \(r\) items from a set of \(n\) items. The binomial coefficient symbol \(\binom{n}{r}\) is preferred when working with multiple combination calculations because it's compact and commonly found on scientific and graphing calculators.

In how many ways can you form a debate team consisting of two coaches and six students if there are five coaches and 18 students interested in joining the team?

To determine the number of ways to put this team together, you use combinations since it doesn't matter in what order the selection of team members is made. You will select two coaches out of five and six students out of eighteen. The formula using the combination is as follows:

• Choosing the coaches:

  The number of ways to choose 2 coaches from 5 is given by:

  \[ \binom{5}{2} \]

• Choosing the students:

  The number of ways to choose 6 students from 18 is given by:

  \[ \binom{18}{6} \]

• Calculating the total combinations:

  Multiply the number of ways to choose the coaches by the number of ways to choose the students:

  \[ \binom{5}{2} \times \binom{18}{6} \]

Each combination term is calculated using the binomial coefficient formula, \(\dbinom{n}{r} = \cfrac{n!}{r!(n-r)!} \), where \(n\) is the total number of items to choose from, and \(r\) is the number of items to choose.

First, we can calculate the number of coach selection combinations:

\(\dbinom{5}{2} = \cfrac{5!}{2!(5-2)!}\)

\(= \cfrac{5!}{2! \times \cancel{3!}}\)

\(= \cfrac{5 \times 4 \times \cancel{3!}}{2 \times 1 \times 3!}\)

\(= \cfrac{20}{2}\)

\(= 10\)

Next, we can calculate the number of student selection combinations :

\(\dbinom{18}{6} = \cfrac{18!}{6!(18-6)!}\)

\(\dbinom{18}{6} = \cfrac{18!}{6!(12)!}\)

\(= \cfrac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times \cancel{12!}}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times \cancel{12!}}\)

\(= \cfrac{18564}{720}\)

\(= 25,740\)

Then, we can calculate the total number of combinations by multiplying the two values:

\(\dbinom{5}{2} \times \dbinom{18}{6} = 10 \times 25,740\)

\(= 257,400\)

Therefore, we can determine there are 257,400 different ways to form such a team.