Question

How many different ways can 5 baseball players and 4 basketball players be selected from 12 baseball players and 13 basketball players

Answer 1

1507 are the different ways can 5 baseball players and 4 basketball players be selected from 12 baseball players and 13 basketball players

Solution:

Given that,

5 baseball players and 4 basketball players be selected from 12 baseball players and 13 basketball players

This is a combination problem

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter

The formula is given as:

$n C_{r}=\frac{n !}{r !(n-r) !}$

Where n represents the total number of items, and r represents the number of items being chosen at a time

Let us first calculate 5 baseball players from 12 baseball players

Here, n = 12 and r = 5

$\begin{array}{l}{12 C_{5}=\frac{12 !}{5 !(12-5) !}} \\\\{12 C_{5}=\frac{12 !}{5 ! \times 7 !}}\end{array}$

For a number n, the factorial of n can be written as:

$n !=n \times(n-1) \times(n-2) \times \ldots . \times 2 \times 1$

Therefore,

$\begin{aligned}12 C_{5} &=\frac{12 \times 11 \times 10 \times \ldots \ldots \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \\\\12 C_{5} &=\frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2} \\\\12 C_{5} &=792\end{aligned}$

Similarly, 4 basketball players be selected 13 basketball players

n = 13 and r = 4

Similarly we get,

$\begin{aligned}&13 C_{4}=\frac{13 !}{4 !(13-4) !}\\\\&13 C_{4}=\frac{13 !}{4 ! \times 9 !}\end{aligned}$

$13C_4 = 715$

Thus total number of ways are:

$12C_5 + 13C_4 = 792 + 715 = 1507$

Thus there are 1507 different ways