Question

A elementary school art class teacher plans to display artwork next to the door of each of the classrooms in the school. Each classroom door will only have one piece of artwork displayed, and the school has 20 such doors. If the teacher has 10 sculptures, 11 sketches, and 9 oil paintings, what is the probability that 4 sculptures, 10 sketches, and 6 oil paintings are chosen to be displayed?

Answer 1

56/8671

Explanation:

First, determine the total number of artworks.

• Sculptures = 10

,

• Sketches = 11

,

• Oil paintings = 9

Total = 10+11+9 = 30

20 artworks can be selected out of 30 in 30C20 ways.

Next:

• 4 sculptures can be selected out of 10 in 10C4 ways.

,

• 10 sketches can be selected out of 11 in 11C10 ways.

,

• 6 oil paintings can be selected out of 9 in 9C6 ways.

The combination formula is:

`^nC_x=(n!)/((n-x)!x!)`

Therefore:

`\begin{gathered} ^(10)C_4=(10!)/((10-4)!4!)=(10!)/(6!4!)=210 \\ ^(11)C_(10)=(11!)/((11-10)!10!)=(11!)/(1!10!)=11 \\ ^9C_6=(9!)/((9-6)!6!)=(9!)/(3!6!)=84 \\ ^(30)C_(20)=(30!)/((30-20)!20!)=(30!)/(10!20!)=30045015 \end{gathered}`

Thus, the probability that 4 sculptures, 10 sketches, and 6 oil paintings are chosen to be displayed is:

`\begin{gathered} (^(10)C_4*^(11)C_(10)*^9C_6)/(^(30)C_(20)) \\ =(210*11*84)/(30045015) \\ =(194040)/(30045015) \\ =(56)/(8671) \end{gathered}`

The probability is 56/8671.