In a box there are 5 black pens, 3 white pens and 4 red pens. In how many different ways can we choose exactly 2 black pens, 2 white pens and 2 red pens from this box?

Introduction / Context:
This is a direct combinations problem involving choosing items of different colours with exact counts from each colour group. The pens of the same colour are considered identical for selection purposes only in terms of colour count; however, since they are simply “pens”, we treat each pen as distinct but constrained by colour. The key idea is to use the combination formula separately for each colour group and then multiply the results because each selection of black pens can be combined with each selection of white and red pens independently.

Given Data / Assumptions:
- Number of black pens: 5.
- Number of white pens: 3.
- Number of red pens: 4.
- Required selection: exactly 2 black pens, 2 white pens, and 2 red pens.
- The order in which we pick pens does not matter; only which pens are in the final group matters.

Concept / Approach:
We handle each colour separately using combinations. From each colour group, we must choose a fixed number of pens. Once we compute the number of ways to choose each colour group, we multiply those counts together by the rule of product because the choices for black, white and red pens are independent of one another. No permutations are needed because we are not arranging the chosen pens in any sequence; we only form a set of six pens obeying the colour constraint.

Step-by-Step Solution:
Step 1: Choose 2 black pens from 5 black pens. The number of ways is 5C2.Step 2: Compute 5C2 = (5 * 4) / (2 * 1) = 10.Step 3: Choose 2 white pens from 3 white pens. The number of ways is 3C2.Step 4: Compute 3C2 = (3 * 2) / (2 * 1) = 3.Step 5: Choose 2 red pens from 4 red pens. The number of ways is 4C2.Step 6: Compute 4C2 = (4 * 3) / (2 * 1) = 6.Step 7: Total number of valid selections = 5C2 * 3C2 * 4C2 = 10 * 3 * 6.Step 8: Compute 10 * 3 = 30, and then 30 * 6 = 180.

Verification / Alternative check:
We can validate by reasoning that any choice of 2 black, 2 white and 2 red pens is uniquely determined by which specific pens were chosen from each colour group. There is no overlap between colours, so there is no double-counting when multiplying the separate combination counts. Re-calculating quickly, 5C2 = 10, 3C2 = 3, 4C2 = 6, and their product 10 * 3 * 6 is clearly 180, which matches the computed result.

Why Other Options Are Wrong:
- 220 and 240 are higher than the correct total and likely originate from miscomputing one or more combinations or accidentally including arrangements where the colour counts are not exactly 2 each.

- 160 is smaller than the correct answer and might result from omitting the red or white selection factor or using 4C2 = 4 instead of 6.

Common Pitfalls:
Some students mistakenly use permutations instead of combinations, multiplying by additional factorials, which would correspond to ordering the selected pens. Others may count only one colour correctly and miscalculate the others. A third common mistake is to try to treat all pens as identical except by colour and then overthink the structure. The straightforward approach is: use combinations colour by colour, then multiply.

Final Answer:
There are 180 different ways to choose exactly 2 black, 2 white and 2 red pens from the box.