Question

There is an unlimited number of red, white, blue and green balls. The balls are all the same except the colour of each ball. What is the number of ways in which selection of 10 balls can be made?

Answer 1

Hint: We will solve the question by the concept of finding integral solutions of an equation of the form a + b +.... + d = n. Here a, b and d are the number of selected objects while n is the total number of selected objects. For example if we take the equation alike a + b + c = 4 then, the solutions for this equation will be given by the formula $C_{r-1}^{n-r+1}$ here n is the total number of selected objects and r is the number of selected objects as here r is 3 which means it consider a, b and c. We will also use the formula $C_{r}^{n}=\dfrac{n!}{\left( n-r \right)!r!}$ in order to solve the question further.

Complete step-by-step answer:
We should first notice here that the total number of balls is infinite. And we need to select the 10 balls from the set containing red, white, blue and green balls. This can be done by collecting all 10 balls of any colour in 10 ways. We take for example that we are selecting 3r, 4w, 2b and 1g making the total of 10 balls. Similarly, we can select any number of pairs here as there is no order.
That is if we suppose that r, w, b and g are respected numbers of balls that are selected by us in which r is the number of red balls, w is the number of white balls, b is the number of blue balls and g is the number of green balls. Then, according to the question the selection is equal to 10 so, we now have r + w + b + g = 10.
Now, if we focus on the equation r + w + b + g = 10 we see that the order here does not matter. Thus, we will not use permutation here. This results in the fact that we will apply a combination. The selection can never be negative. Therefore, we are actually considering the numbers greater than 0. Thus, we have r + w + b + g $\ge $ 0 and r + w + b + g = 10. So, the number of non-negative integral solutions of this equation is equal to the number of ways in which we can distribute n = 10 identical balls into r = 4 different balls. The formula for this is given by $C_{r-1}^{n+r-1}$. Thus, we get
 $\begin{align}
  & C_{r-1}^{n+r-1}=C_{4-1}^{10+4-1} \\
 & \Rightarrow C_{r-1}^{n+r-1}=C_{3}^{14-1} \\
 & \Rightarrow C_{r-1}^{n+r-1}=C_{3}^{13} \\
\end{align}$
By the formula $C_{r}^{n}=\dfrac{n!}{\left( n-r \right)!r!}$ we will get
$\begin{align}
  & C_{3}^{13}=\dfrac{13!}{\left( 13-3 \right)!3!} \\
 & \Rightarrow C_{3}^{13}=\dfrac{13!}{10!3!} \\
 & \Rightarrow C_{3}^{13}=\dfrac{13\times 12\times 11\times 10!}{10!3!} \\
 & \Rightarrow C_{3}^{13}=\dfrac{13\times 12\times 11}{3!} \\
 & \Rightarrow C_{3}^{13}=\dfrac{13\times 12\times 11}{3\times 2\times 1} \\
 & \Rightarrow C_{3}^{13}=\dfrac{13\times 2\times 11}{1} \\
 & \Rightarrow C_{3}^{13}=13\times 22 \\
 & \Rightarrow C_{3}^{13}=286 \\
\end{align}$
Hence, the required number of ways is 286.

Note: While solving the question one gets confused about what type of permutation combination is going to use here. We cannot solve it by using permutations because order here does not matter. We will use a combination then. But the combination formula that we are using here is followed by an equation. This is because we will use this combination for an infinite number of objects. Whenever we come across such types of unlimited supplies then we will solve it only by the formula $C_{r-1}^{n+r-1}$. This is the formula that is used in finding the non negative integral solutions. This formula is different from the formula where there is a need to find the solutions for only positive numbers of solutions of an equation.