Question

There are 21 balls which are either black or white and the balls of the same color are alike. Find the number of white balls so that the number of arrangements of these balls in row is maximum.

Answer 1

Hint: In this question, the total number of balls is given and we are asked to find the number of white balls to make the number of arrangements of these balls in row maximum. Therefore, we should use the theory of permutation and combination to find the number of arrangements when there are w white balls and then find w such that the expression for the number of arrangements becomes maximum.
Complete step-by-step answer:
Let the number of white balls be equal to w……………………...(1.1)
It is given that the total number of balls is equal to 21. Therefore, the number of black balls should be the total number of balls minus the number of white balls, therefore the number of black balls should be (21-w)…………………………(1.2)
Now, form the theory of permutation, the number of arrangements of total n objects, with k distinct type of objects, out of which ${{r}_{1}}$ are of the first type, ${{r}_{2}}$ are of the second type and so on and balls of each type are alike is given by
$\text{no}\text{. of arrangements=}\dfrac{n!}{{{r}_{1}}!{{r}_{2}}!...{{r}_{k}}!}.........(1.3)$
In this case, the total number of objects=n=21
There are two types of balls, therefore k=2, and from (1.1) and (1.2), we have
No. of white balls=${{r}_{1}}$=w
No. of black balls=${{r}_{2}}$=21-w
Therefore, using these values in equation (1.3), we get
$\text{no}\text{. of arrangements=}\dfrac{21!}{w!\left( 21-w \right)!}.........(1.4)$
However, we know that the combination of 21 objects taken w at a time is also given by
${}^{21}{{C}_{w}}=\dfrac{21!}{w!\left( 21-w \right)!}$
Comparing it with equation (1.4), we find that
$\text{no}\text{. of arrangements=}{}^{21}{{C}_{w}}...............(1.5)$
Also, we know that if n is odd, then ${}^{n}{{C}_{r}}$ is maximum if $r=\left( \dfrac{n-1}{2} \right)$ or $r=\left( \dfrac{n+1}{2} \right)$. Therefore, using this in equation (1.5), we find that the number of arrangements is maximum if $w=\left( \dfrac{21-1}{2} \right)=10$ or $w=\left( \dfrac{21+1}{2} \right)=11$.
Therefore, the required number of white balls is 10 or 11 which is the required answer to this question.

Note: In equation (1.3), we should note that the division of ${{r}_{1}}!{{r}_{2}}!...{{r}_{k}}!$ holds only if objects of each type are alike and thus indistinguishable. Therefore, if it was not given that the black and white balls are to be considered alike, we should have divided $21!\left( 21-w \right)!$ in equation (1.4).