Ask Question
16 February, 02:38

Members of a hospital advisory committee will be selected from 5 doctors and 4 nurses. The committee must have 2 doctors and 2 nurses.

How many possible combinations of doctors and nurses could be chosen for the committee?

+4
Answers (1)
  1. 16 February, 03:24
    0
    16 possible ways

    Step-by-step explanation:

    Using the formular Cp, k = p!/I! (P-k) !

    Where k = number of selected committee members

    P = available number that can sit in the committee

    For doctors, there are 5 doctors

    C = 5!/2! (3) ! = 120/12 = 10ways

    For nurses, there are 4

    C = 4!/2! (2) ! = 24/4 = 6ways

    Total possible ways=10 + 6 = 16 ways
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Members of a hospital advisory committee will be selected from 5 doctors and 4 nurses. The committee must have 2 doctors and 2 nurses. How ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers