An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which

Q: Why do we group Black and White marbles together? Doesn't that lose information?

That is a brilliant observation! In combinatorics, we only care about the properties that affect our constraints. Since the problem only restricts the number of Red marbles, the distinction between Black and White is irrelevant to the count. By grouping them into 'Non-Red', we simplify our mental model without losing any mathematical validity. It is the art of focusing only on what matters.

Q: Is the complementary method always faster?

Often, yes! In JEE Advanced, time is your most precious resource. Whenever you see 'at most' or 'at least', always check if the complementary case (the 'forbidden' cases) is smaller or easier to calculate than the valid cases. If there is only one forbidden case, as in this problem, the complementary method is a massive time-saver.

Q: Wait, why do we use combinations ($^{n}C_{r}$) instead of permutations ($^{n}P_{r}$)?

Remember, the problem asks for the number of ways to 'draw' marbles. When you draw a handful of marbles from an urn, the order in which they land in your hand does not matter—you just care about the final set you are holding. Whenever order is irrelevant, we use combinations. If the problem had asked for the order of drawing, we would have used permutations.

Question

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at most three of them are red is __________.

Sigma Percentile · Accepted Answer

### The Urn of Possibilities: A Combinatorial Journey

Imagine you are standing before an urn. It is not just a container; it is a universe of possibilities. Inside, you have $5$ Red, $4$ Black, and $3$ White marbles.

Your task is to reach in and pull out $4$ marbles. There is a constraint that defines the rules of this game: you must have **at most three red marbles**.

In the high-stakes arena of JEE Advanced, problems like this are not just about calculation; they are about strategy. Let us break it down.

### Phase 1: The Art of Simplification

The first step to solving any complex problem is to strip away the noise. We have three colors, but the constraint only cares about the Red ones.

The Black and White marbles are essentially "not Red." So, let us simplify our world: we have $5$ Red marbles and $4 + 3 = 7$ Non-Red marbles.

By grouping the Black and White marbles, we have reduced a three-variable problem into a binary one. This is the first mark of a master problem solver: simplifying the system to its core essence.

### Phase 2: The Path of Summation

Now, let us look at the condition: "at most three red." This means we are satisfied with $0$, $1$, $2$, or $3$ red marbles. We can calculate the number of ways for each scenario and sum them up.

For each case, we use the Fundamental Principle of Counting. If we want $k$ Red marbles, we must choose $k$ from the $5$ available, and then choose the remaining $4-k$ marbles from the $7$ Non-Red ones.

1. **Zero Red Marbles:** We choose $0$ from $5$ Red and $4$ from $7$ Non-Red.
$$^{5}C_{0} 	imes ^{7}C_{4} = 1 	imes 35 = 35 	ext{ ways}$$

2. **Exactly One Red Marble:** We choose $1$ from $5$ Red and $3$ from $7$ Non-Red.
$$^{5}C_{1} 	imes ^{7}C_{3} = 5 	imes 35 = 175 	ext{ ways}$$

3. **Exactly Two Red Marbles:** We choose $2$ from $5$ Red and $2$ from $7$ Non-Red.
$$^{5}C_{2} 	imes ^{7}C_{2} = 10 	imes 21 = 210 	ext{ ways}$$

4. **Exactly Three Red Marbles:** We choose $3$ from $5$ Red and $1$ from $7$ Non-Red.
$$^{5}C_{3} 	imes ^{7}C_{1} = 10 	imes 7 = 70 	ext{ ways}$$

Summing these up: $35 + 175 + 210 + 70 = 490$. We have arrived at the answer!

### Phase 3: The Elegant Shortcut

As an elite student, you should always look for the "elegant" path. We can avoid calculating four separate cases by using the **Complementary Method**.

Think about the total number of ways to draw any $4$ marbles from the $12$ available. That is $^{12}C_{4}$. This represents every possible combination, regardless of color.

$$^{12}C_{4} = \frac{12 	imes 11 	imes 10 	imes 9}{4 	imes 3 	imes 2 	imes 1} = 495$$

The only scenario we are forbidden from having is drawing $4$ Red marbles. Calculating this forbidden case is trivial:
$$^{5}C_{4} = 5$$

Now, subtract the forbidden cases from the total:

$$Total 	ext{ } Ways - Forbidden 	ext{ } Ways = 495 - 5 = 490$$

Look at that! In one line of logic, we reached the same result. This is the power of mathematical maturity. Whether you choose the path of summation or the path of subtraction, you are building the intuition that will carry you through the toughest exams.

Animated Solution for Mathematics - Permutations and Combinations: An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at most three of them are red is __________.

Enter Numerical Value:

Visualized Solution

Analyze the Urn Contents

Grouping by Condition

Interpret 'At Most 3 Red'

Case 1: Zero Red Marbles

Compute Case 1

Case 2: Exactly One Red

Compute Case 2

Case 3: Exactly Two Red

Compute Case 3

Case 4: Exactly Three Red

Compute Case 4

Summing the Total Ways

Pro Tip: Complementary Method

The Sigma Insight: Combinations and Selection

The Urn of Possibilities

Phase 1

Phase 2

Phase 3

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