Question

Two new drugs are to be tested using a group of 60 laboratory mice, each tagged with a number for identification purposes. Drug A is to be given to 22 mice, drug B is to be given to another 22 mice, and the remaining 16 mice are to be used as controls. How many ways can the assignment of treatments to mice be made? (A single assignment involves specifying the treatment for each mouse—whether drug A, drug B, or no drug.) (Enter the exact number or an equivalent algebraic expression.)

Answer 1

Answer:

$\frac{60!}{22!22!16!}$

Step-by-step explanation:

As given, drug A is to be given to 22 mice, drug B is to be given to another 22 mice, and the remaining 16 mice are to be used as controls.

Required number of ways = Number of ways to select mice that gets drug A x the number of ways for mice that gets drug B x the number of ways the mice gets no drugs.

= $\frac{60!}{22!38!} \times \frac{38!}{22!16!} \times1$

Solving this we get;

= $\frac{60!}{22!22!16!}$

= 314,790,828,599,338,321,972,833,000