# Probability that all patients get the right medicine

One nurse missed the doctor's instructions with the medication she has to give to $$10$$ patients. $$4$$ patients are taking a pill for hypertension, $$3$$ patients are taking a pill for diabetes, $$2$$ patients are taking a pill for heart arrhythmia and $$1$$ patient is taking a pill for hypotension. The nurse randomly gives the pills to the patients. Coincidentally, everyone receives medication in the right way. What is the probability that this will happen?

I think that all the cases are $$\dfrac{10!}{4! \cdot 3! \cdot 2! \cdot 1!}$$ but I face difficulty in finding the sum of cases that each patient takes the right medicine.I would appreciate for your quidance. Thank you very much in advance.

• Consider permutations of $(a,a,b)$ amongst three patients. The number of cases is $\frac{3!}{2!}=3$: $\{(a,a,b),(a,b,a),(b,a,a)\}$. How many are correct? is it not just 1? – Rahul Madhavan Apr 17 at 17:27
• You helped me a lot, Thank you very much! – Magda Apr 17 at 19:34

Suppose the patients are given the pills in the order $$H_1H_2H_3H_4D_1D_2D_3A_1A_2O_1$$ where $$H_i$$ represents the person who has to take the pill for hypertension, $$D_i$$ for diabetes, $$A_i$$ for arrhythmia and $$O_1$$ for hypotension.
Now, probability that $$H_1$$ gets the right pill is $$4/10$$, then $$H_2$$ gets the right pill is $$3/9$$ so in this way, the probability that all get the right pill would be $$\dfrac{4}{10} \cdot \dfrac{3}{9} \cdot \dfrac{2}{8} \cdot \dfrac{1}{7} \cdot \dfrac{3}{6} \cdot \dfrac{2}{5} \cdot \dfrac{1}{4} \cdot \dfrac23 \cdot \dfrac12 \cdot 1$$