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)=(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)=(12 !)/(5 !(12-5) !)} \\\\{12 C_(5)=(12 !)/(5 ! * 7 !)}\end{array}`

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

`n !=n *(n-1) *(n-2) * \ldots . * 2 * 1`

Therefore,

`\begin{aligned}12 C_(5) &=(12 * 11 * 10 * \ldots \ldots * 2 * 1)/(5 * 4 * 3 * 2 * 1 * 7 * 6 * 5 * 4 * 3 * 2 * 1) \\\\12 C_(5) &=(12 * 11 * 10 * 9 * 8)/(5 * 4 * 3 * 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)=(13 !)/(4 !(13-4) !)\\\\&13 C_(4)=(13 !)/(4 ! * 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