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 21 mice, drug B is to be given to another 21 mice, and the remaining 18 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:

4.979044478499338 × 10²⁶

Step-by-step explanation:

Combination can be used to determine the number of ways the mice can be selected for the drugs (A, B) and the control group.

Combination factorial is define by  ⁿCr = $\frac{n!}{(n-r)!r!}$

21 group of mice receiving Drug A  can be selected in ⁶⁰C₂₁ = $\frac{60!}{21!39!}$

(60 - 21 = 39 ) mice remained for selection of 21 mice for the second drug

Drug B 21 mice can be chosen with ³⁹C₂₁ = $\frac{39!}{21!18!}$

( 39 - 21 = 18)  remained for control with ¹⁸C₁₈ = $\frac{18!}{18!0!}$

The number of ways the mice can be chosen for drug A, drug B and the control = ⁶⁰C₂₁ × ³⁹C₂₁ × ¹⁸C₁₈ = $\frac{60!}{21!39!}$ × $\frac{39!}{21!18!}$ × $\frac{18!}{18!0!}$ = 4.979044478499338 × 10²⁶