There are $5$ boxes numbered from $1$ to $5$. There is $1$ Red and $2k$ black balls in the $k^{th}$ box

combinationscombinatorics

There are $5$ boxes numbered from $1$ to $5$. There is $1$ Red and $2k$ black balls in the $k^{th}$ 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
colours are all alike).

Now which of the following holds good?

(A) Total number of ways selecting odd number of red balls is $4725$

(B) Total number of ways of selecting even number of red balls is $5670$.

(C) Total number of ways selecting odd number of red balls is $5670$

(D) Total number of ways of selecting even number of red balls is $4725$

I cannot think of how to proceed this problem

Best Answer

Strategy:

  1. Selecting an odd number of red balls: This means you select the only red ball from one, three, or five of the boxes. From the other boxes, you select from up to $2k$ black balls, but you must choose at least one.
  2. Selecting an even number of red balls: This means you select the only red ball from zero, two, or four of the boxes. From the other boxes, you select from up to $2k$ black balls, but you must choose at least one.

To illustrate with a particular case. Suppose we select red balls from the first three boxes. There is one possible selection from the first box (the red ball), one possible selection from the second box (the red ball), one possible selection from the third box (the red ball), eight possible selections from the fourth box (since we can choose $1, 2, 3, 4, 5, 6, 7,$ or $8$ black balls), and $10$ selections from the fifth box (since we can choose $1, 2, 3, 4, 5, 6, 7, 8, 9,$ or $10$) of the black balls. Therefore, this case corresponds to $1 \cdot 1 \cdot 1 \cdot 8 \cdot 10 = 80$ possible choices.

Notice that there are $\binom{5}{3} = 10$ ways to select from which three of the five boxes you will select a red ball. In each case, the number of black balls you can choose depends on which boxes remain. We considered one of these $10$ cases above.

There are $\binom{5}{j}$ cases to consider when you select exactly $j$ red balls.

Can you proceed?

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