There are 5 boxes numbered from 1 to 5. There is 1 Red and $2 k$ black balls in k^th box, $k = 1, 2, 3, 4, 5 .$ From each box either one red ball is taken or one or more than one black balls are taken. But from each box both coloured balls are never taken (ball of same colour are all alike). Now which of the following holds good ?

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Total number of ways selecting odd number of red balls is 4725

Total number of ways of selecting even number of red balls is 5670

Total number of ways selecting odd number of red balls is 5670

Total number of ways selecting even number of red balls is 4725

answer is A, B.

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Detailed Solution

Let
x : Total number of ways of selecting even number of red balls
y : Total number of ways of selecting odd numbers of red balls
(2 – 1) (4 – 1) (6 – 1) (8 – 1) (10 – 1) = x – y
(2 + 1) (4 + 1) (6 + 1) (8 + 1) (10 + 1) = x + y
$\begin{array}{l} \Rightarrow x + y = 10395 \\ x - y = 945 \end{array}$
$\Rightarrow x = 5670, y = 4725$