Question

An art collector who owns 10 original paintings is preparing a will. In how many ways could she leave 4 of the paintings to her daughter?

Answer 1

The art collector has 10 painting.

He will leave 4 to his doughter.

To calculate the number of ways he coud leave those 4 paintings we use the following for combinations without repetition:

`\text{Combinations}=(n!)/(r!(n-r)!)`

Where n is the total number of painting: n=10

r is the amount he is going to leave: r=4

Thus we get the following result for the combinations that he can make:

`\begin{gathered} \text{Combinations}=(10!)/(4!(10-4)!) \\ \text{Combinations}=(10!)/(4!(6)!) \\ \text{Combinations}=210 \end{gathered}`

Answer:

In how many ways could she leave 4 of the paintings to her daughter?​ 210