There are 21 balls which are either white or black and the balls of th

Question

Competition Mode Question

There are 21 balls which are either white or black and the balls of the same colour are alike. If the number of arrangements of these balls in a row be maximum then the number of white balls can be

A. 10

B. 12

C. 14

D. 20

E. 21

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CareerGeek · Answer

Let the number of white balls =  w 
So, number of black balls =  21 - w 
Number of ways of arranging w identical white balls & (21 - w) identical black balls in a straight line =  \frac{21!}{w!*(21 - w)!} = 21c_w

A. w = 10 --> Number of arrangements possible =  21c_{10} 
B. w = 12 --> Number of arrangements possible =  21c_{12} = 21c_9 
C. w = 14 --> Number of arrangements possible =  21c_{14} = 21c_7 
D. w = 20 --> Number of arrangements possible =  21c_{20} = 21c_1 
E. w = 21 --> Number of arrangements possible =  21c_{21} = 1

So, maximum number of arrangements possible = highest value among the options =  21c_{10}

Option A

freedom128 · Answer

21 = white + black

Objective: the number of arrangements of these balls in a row must be maximum

If no of white balls is:
 A. 10 --> 21C10= 352,716 ways  
B. 12 --> 21C12= 293,930 ways
C. 14 --> 21C14= 116,280 ways
D. 20 --> 21C20= 21 ways
E. 21 --> 21C21= 1 way

Clearly, Final Answer is (A)

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shameekv1989 · Answer

There are 21 balls which are either white or black and the balls of the same colour are alike. If the number of arrangements of these balls in a row be maximum then the number of white balls can be

A. 10

B. 12

C. 14

D. 20

E. 21

Since the total number of balls are odd and the same color balls are alike (same in number) 
The only combination that fits would be 1+20 which would mean 1 Black ball and 20 White balls

Therefore maximum number is 20 White balls - Answer - D

nick1816 · Answer

1. When n is even
nCr is max if  r=\frac{n}{2}

2. When n is odd
nCr is max, if  r=\frac{n±1}{2}

Suppose we have 'r' number of White balls.

Total possible arrangements=  \frac{21!}{r!(21-r)!}  = 21Cr

21Cr will be max, when r is equal to  \frac{21±1}{2}  = 10 or 11

A

exc4libur · Answer

If we have x white we will have 21!-x black;
and the no of arrangements will be 21!/x!*(21-x)!

(E) if w=21, then no ways: 21!/(21-21)!*21!=21!/1*21!=1
(D) if w=20, then no ways: 21!/(21-20)!*20!=21!/1*20!=21

Notice that the smaller the no of w, the larger the no of ways

Ans (A)

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