Question

At a blood drive, 6 donors with type O + blood, 4 donors with type A + blood, and 2 donors with type B + blood are in line. In how many distinguishable ways can thedonors be in line?The donors can be in line in different ways.2

Answer 1

First, let's state the given information:

The number of donors with the blood of type O+ is 6.

The number of donors with the blood of type A+ is 4.

The number of donors with the blood of type B+ is 2.

we will label these numbers as follows for reference:

`\begin{gathered} a=6 \\ b=4 \\ c=2 \end{gathered}`

We will also need the total number of donors "n":

`\begin{gathered} n=6+4+2 \\ n=12 \end{gathered}`

Since we are asked to find the distinguishable ways in which the donors can be in line, we need to find the number of permutations.

The formula for permutations is:

`P=(n!)/(a!b!c!)`

Where "!" means factorial, and is defined as follows, for example, 4! is:

`4!=4*3*2*1`

And so on depending on the number.

In this case, the operation we have to solve is:

`P=(12!)/(6!4!2!)`

Solving this division we get the result for the number of distinguishable ways in which the donors can be in line:

`P=13,860`

Answer: 13,860