Respuesta :
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:
[tex]n C_{r}=\frac{n !}{r !(n-r) !}[/tex]
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
[tex]\begin{array}{l}{12 C_{5}=\frac{12 !}{5 !(12-5) !}} \\\\{12 C_{5}=\frac{12 !}{5 ! \times 7 !}}\end{array}[/tex]
For a number n, the factorial of n can be written as:
[tex]n !=n \times(n-1) \times(n-2) \times \ldots . \times 2 \times 1[/tex]
Therefore,
[tex]\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}[/tex]
Similarly, 4 basketball players be selected 13 basketball players
n = 13 and r = 4
Similarly we get,
[tex]\begin{aligned}&13 C_{4}=\frac{13 !}{4 !(13-4) !}\\\\&13 C_{4}=\frac{13 !}{4 ! \times 9 !}\end{aligned}[/tex]
[tex]13C_4 = 715[/tex]
Thus total number of ways are:
[tex]12C_5 + 13C_4 = 792 + 715 = 1507[/tex]
Thus there are 1507 different ways
