Permutations/ Combinations- Dividing a pile into 3 piles

Question

I have been given the following question:

---

A bag contains ten balls numbered from one to ten.

• How many distinct combinations can be formed by drawing three balls from the bag? (The order in which the balls are drawn is irrelevant.)
• In how many distinct ways can the balls be divided among three piles containing five, three and two balls respectively?

---

My attempt at a solution

---

• The permutation relation for this problem can be expressed: $$ P(N,k)=\frac{N!}{k!(N-k)!},$$ Where $N$ is the total number of balls $\left(10\right)$, $k$ is the number of balls drawn, hence by substitution, we see: $$P(10,3)=120,$$ This is the number of permutations, but we have been told the order in which the balls are drawn is irrelevant, so we must consider the combination relation: $$C(N,k)=\frac{P(N,k)}{k!},$$ Hence: $$C(10,3)=\frac{120}{6}=20.$$
• My problem is with the second part, I am thinking that I can visualize this, in the same way, I did the first part, as extracting five, three and two balls from the 10, so, using: $$ P(N,k)=\frac{N!}{k!(N-k)!},$$ $$ P(10,5)=\frac{10!}{5!(10-5)!}=252,$$ $$ P(10,3)=\frac{10!}{3!(10-3)!}=120,$$ $$ P(10,2)=\frac{10!}{2!(10-2)!}=45,$$ $$\text{Total permutations}= P(10,5)+P(10,3)+P(10,2)=417.$$ My problem is, I am unsure whether I should consider the total amount of balls to reduce after the first pile is created, i.e.: $$ P(10,5)=\frac{10!}{5!(10-5)!}=252,$$ $$ P(5,3)=\frac{5!}{3!(5-3)!}=10,$$ $$ P(2,2)=\frac{2!}{2!(2-2)!}=1,$$ $$\text{Total permutations}= P(10,5)+P(5,3)+P(2,2)=263.$$ At first, I thought it seemed logical to consider the second method but I think I am over complicating the problem, as it doesn't state the order that the piles are created in. If someone could provide some clarification of the problem, that'd be great, as I am currently abit confused. Thanks.

Answer 1

Your second method is correct.

In the first method you are assuming you can use the same balls in more than one piles which is clearly wrong because one ball can only be in one pile at a time.

Also when doing the second method you will have to multiply instead of adding because you are performing the operations one after the another.

Answer 2

You are using P for the 2nd part when you should be using C. Although you have calculated combinations. Also you've added when you should multiply. For every one of the 252 combinations of 5 balls, there are 10 combinations of the other 5 balls.

$$^{10}C_5\cdot ^5C_3 \cdot ^2C_2 = 2520$$