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Question: 2m white counters and 2n red counters are arranged in a straight line with (m+n) counters on each si...

2m white counters and 2n red counters are arranged in a straight line with (m+n) counters on each side of a central mark. The number of ways of arranging the counters, so that the arrangements are symmetrical with respect to the central mark, is
(a) m+nCm{}^{m+n}{{C}_{m}}
(b) 2m+2nC2m{}^{2m+2n}{{C}_{2m}}
(c) 12 (m+n)!m!n!\dfrac{1}{2}\text{ }\dfrac{(m+n)!}{m!n!}
(d) None of them

Explanation

Solution

Firstly, we are given a number of white and red counters that should be equal on both the left and right side of the central mark. Then, the number of counters will be half on both sides. Thus, one side contains (m+n) counters. Then, we get the number of ways to arrange the counters symmetrically.

Complete step-by-step answer:
We have been given 2m white counters and 2n red counters arranged in a straight line with (m+n) counters on each side of the central mark.
We need to find the number of ways to arrange these counters symmetrically.
Now there's a central point, the no. of white counters on the left and right side of the counters should be equal. Similarly, the no. of red counters on the left and right side should be equal.
Hence we have half of 2m will be on the left and the other half on the right.
Similarly, we also have half of 2n will be on the right and the other half on the right.
__ __ __ __ __ __ __ __ •__ __ __ __ __ __ __ __

            m + n m + n

Hence, there are m white and n red on the left side and m white and n red on the right side and we can arrange white and red counters in any order.
Now, let us take white as W and red as R and we can arrange it as:
WWRWRWR•RWRWRWW
If we arrange one side, the same gets arranged on the other side, as it's symmetrical. It is more like a minor image. You can understand the same from the above.
Hence we can m white and n red in (m+n) places as (m+n)! and, m and n are identical. Thus divide (m+n)! by m! and n!
So, number of ways of arrangement=(m+n)!m!n!=\dfrac{(m+n)!}{m!n!}
Then, we can also write the above as m+nCm{}^{m+n}{{C}_{m}}or m+nCn{}^{m+n}{{C}_{n}}as combination because order is not important.
So, we know by the concept of combination asnCr = n!(n−r)! r!{}^{n}{{C}_{r}}\text{ = }\dfrac{n!}{(n-r)!\text{ r!}}.
m+nCm = (m+n)!(m+n−m)! m!  ⇒m+nCm = (m+n)!n! m! \begin{aligned} & {}^{m+n}{{C}_{m}}\text{ = }\dfrac{(m+n)!}{(m+n-m)!\text{ m!}}\text{ } \\\ & \Rightarrow {}^{m+n}{{C}_{m}}\text{ = }\dfrac{(m+n)!}{n!\text{ m!}} \\\ \end{aligned}
m+nCn = (m+n)!(m+n−n)! n!  ⇒m+nCn= (m+n)!n! m! \begin{aligned} & {}^{m+n}{{C}_{n}}\text{ = }\dfrac{(m+n)!}{(m+n-n)!\text{ n!}}\text{ } \\\ & \Rightarrow {}^{m+n}{{C}_{n}}\text{= }\dfrac{(m+n)!}{n!\text{ m!}} \\\ \end{aligned}
Thus we can say that,
(m+n)!m!n!=m+nCm=m+nCn\dfrac{(m+n)!}{m!n!}\text{=}{}^{m+n}{{C}_{m}}\text{=}{}^{m+n}{{C}_{n}}
So, the counters can be arranged in m+nCn{}^{m+n}{{C}_{n}}ways symmetrically with respect to the central mark.

So, the correct answer is “Option A”.

Note: It is said that both sides are symmetric, so don't get confused. It's more like taking a mirror image of the left side. You might probably get the answer as (m+n)!m!n!\dfrac{(m+n)!}{m!n!}, which does not belong to any option. Hence remember to convert it into combination form as, nCr{}^{n}{{C}_{r}} form.