JEE advance 2022 paper 2 ball selection

Question:

Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

Solution 1:

We start by noting that each of the 4 boxes contains 3 red and 2 blue distinct balls. In any valid selection of 10 balls, from each box at least one red and one blue must be chosen.

Step 1. Determine How Many Balls Are Chosen from Each Box

Let $x_i$ be the number of balls chosen from box $i$ (for $i=1,2,3,4$). We have:

$$x_1+x_2+x_3+x_4 = 10$$

Since from each box at least one red and one blue must be chosen, we must choose at least 2 balls from each box:

$$x_i \ge 2 \quad \text{for } i=1,2,3,4.$$

Define $y_i = x_i - 2$ (so $y_i \ge 0$). Then:

$$y_1+y_2+y_3+y_4 = 10 - 8 = 2.$$

The number of nonnegative integer solutions of this equation (by stars and bars) is

$$\binom{2+4-1}{4-1} = \binom{5}{3} = 10.$$

However, rather than simply counting the solutions, we need to consider the two types of distributions.

Step 2. Classify the Distributions

Since $y_1+y_2+y_3+y_4 = 2$, there are two types:

1. Type A: One box gets 2 extra balls and the others get 0 extra.
   In this case, one box has $x = 2+2 = 4$ balls and each of the other three boxes has $x = 2$ balls. There are 4 ways to choose which box gets 4 balls.

2. Type B: Two boxes get 1 extra ball each and the others get 0 extra.
   Then two boxes have $x = 2+1 = 3$ balls each and the other two boxes have $x = 2$ balls each. There are

   $$\binom{4}{2} = 6$$

   ways to choose which two boxes get 3 balls.

Step 3. Count Valid Selections Within Each Box

For any box, it contains 3 red and 2 blue balls (all distinct). Let $f(n)$ be the number of ways to choose $n$ balls from a box so that there is at least one red and one blue.

• For $n=2$:
  Total ways to choose 2 from 5 is $\binom{5}{2}=10$.
  All-red selections: $\binom{3}{2}=3$.
  All-blue selections: $\binom{2}{2}=1$.
  Thus,

  $$f(2)=10-3-1=6.$$

• For $n=3$:
  Total ways: $\binom{5}{3}=10$.
  All-red selections: $\binom{3}{3}=1$.
  (All-blue selections are impossible since there are only 2 blue.)
  Thus,

  $$f(3)=10-1=9.$$

• For $n=4$:
  Total ways: $\binom{5}{4}=5$.
  All-red and all-blue selections are impossible because choosing 4 red is not possible (only 3 available) and 4 blue is not possible (only 2 available).
  Thus,

  $$f(4)=5.$$

Step 4. Combine the Two Types

Type A Distribution $(4,2,2,2)$:

• For the box with 4 balls: $f(4)=5$.
• For each of the 3 boxes with 2 balls: $f(2)=6$ each.
• Total ways for a fixed distribution: $$5 \times 6^3 = 5 \times 216 = 1080.$$
• There are 4 choices for which box is the “4-ball” box, so total for Type A: $$4 \times 1080 = 4320.$$

Type B Distribution $(3,3,2,2)$:

• For each box with 3 balls: $f(3)=9$ (2 boxes).
• For each box with 2 balls: $f(2)=6$ (2 boxes).
• Total ways for a fixed distribution: $$9^2 \times 6^2 = 81 \times 36 = 2916.$$
• There are 6 ways to choose which boxes get 3 balls, so total for Type B: $$6 \times 2916 = 17496.$$

Step 5. Final Count

Add the two types:

$$4320 + 17496 = 21816.$$

Thus, the total number of different ways to choose 10 balls from the 4 boxes, ensuring that from each box at least one red and one blue ball are chosen, is 21816.




Solution 2:

Below is a step‐by‐step way to see how the alternate solution is constructed. It is essentially the same reasoning as the “two‐case” method (one box contributes 4 balls vs. two boxes contribute 3 balls each), but it uses combinations directly for each box:

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Overview of the Problem

• We have 4 boxes, each containing 3 red (R) and 2 blue (B) distinct balls.
• We want to choose a total of 10 balls, with the requirement that from each box we pick at least one red and one blue.

We split the counting into two main Cases based on how many balls we take from each box:

1. Case I: $(4,2,2,2)$ distribution — One box contributes 4 balls; each of the other three boxes contributes 2 balls.
2. Case II: $(3,3,2,2)$ distribution — Two boxes contribute 3 balls each; the other two boxes contribute 2 balls each.

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Key Combinatorial Counts Per Box

Since each box has 3 reds and 2 blues, the solution repeatedly uses these small “building blocks”:

1. Number of ways to choose 2 balls (with ≥1R, ≥1B):

   $$\binom{2}{1}\cdot \binom{3}{1} \;=\; 2 \times 3 \;=\; 6.$$

   Explanation: you must pick exactly 1 blue out of the 2 available, and 1 red out of the 3 available.

2. Number of ways to choose 3 balls (with ≥1R, ≥1B):
   We can break it into sub‐cases:

   • Pick 1 blue + 2 reds: $\binom{2}{1}\cdot \binom{3}{2} = 2 \times 3 = 6$.
   • Pick 2 blues + 1 red: $\binom{2}{2}\cdot \binom{3}{1} = 1 \times 3 = 3$.
     Adding them gives $6 + 3 = 9.$

3. Number of ways to choose 4 balls (with ≥1R, ≥1B):
   This can be counted by “which single ball is not chosen,” or directly as:

   $$\binom{2}{1} + \binom{3}{1} \;=\; 2 + 3 \;=\; 5.$$
   • Omit 1 blue (2 ways) ⇒ keep all 3 reds + the other blue.
   • Omit 1 red (3 ways) ⇒ keep the other 2 reds + both blues.

So in shorthand:

• $f(2) = 6$,
• $f(3) = 9$,
• $f(4) = 5$.

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Case I: $(4,2,2,2)$

1. Choose which 1 box will contribute 4 balls: $\binom{4}{1} = 4$.
2. From that chosen box, pick 4 (with ≥1R, ≥1B): $f(4) = 5.$
3. From each of the remaining 3 boxes, pick 2 (with ≥1R, ≥1B): $f(2) = 6$ for each box.

Putting it together:

$$\underbrace{\binom{4}{1}}_{\text{which box for 4 balls}} \times \underbrace{f(4)}_{\text{ways to pick 4}} \times \underbrace{f(2)^3}_{\text{ways to pick 2 from each of 3 boxes}} \;=\; 4 \times 5 \times 6^3 \;=\; 4 \times 5 \times 216 \;=\; 4320.$$

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Case II: $(3,3,2,2)$

1. Choose which 2 boxes will contribute 3 balls each: $\binom{4}{2} = 6.$
2. From each of those 2 boxes, pick 3 (with ≥1R, ≥1B): $f(3) = 9$ per box.
3. From the remaining 2 boxes, pick 2 (with ≥1R, ≥1B): $f(2) = 6$ per box.

Hence:

$$\underbrace{\binom{4}{2}}_{\text{which 2 boxes for 3 balls}} \times \underbrace{f(3)^2}_{\text{ways to pick 3 from each}} \times \underbrace{f(2)^2}_{\text{ways to pick 2 from each}} \;=\; 6 \times 9^2 \times 6^2 \;=\; 6 \times 81 \times 36 \;=\; 17496.$$

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Total Number of Ways

Summing the two cases:

$$4320 \;+\; 17496 \;=\; 21816.$$

This matches the result we get by any other method.

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Why the Expressions Look Like $\binom{2}{1} + \binom{3}{1}$, $\binom{2}{1}\binom{3}{1}$, etc.

• When picking 4 from the box, the expression $\binom{2}{1} + \binom{3}{1}$ is simply counting how many ways to omit exactly one ball: either omit one of the 2 blues or one of the 3 reds.
• When picking 2 from the box, the product $\binom{2}{1}\binom{3}{1}$ counts how to pick exactly 1 blue (out of 2) and 1 red (out of 3).
• When picking 3 from the box, the sum $\binom{2}{1}\binom{3}{2} + \binom{2}{2}\binom{3}{1}$ counts the two ways to ensure at least 1 red and 1 blue: either (1B + 2R) or (2B + 1R).

All these small building blocks then get multiplied according to which boxes we take 2, 3, or 4 balls from, and finally we sum over the different ways to assign those numbers of balls to the boxes.