Six X's have to be placed in the squares of figure below in such a way that each row contains at lea

Q: Wait, why can't we just use the standard Stars and Bars formula for $x_1 + x_2 + x_3 = 6$?

That is a fantastic question! The standard Stars and Bars formula, $\binom{n+k-1}{k-1}$, assumes that each bin has an infinite capacity. In our problem, the rows have strict physical limits: $x_1 \le 2$, $x_2 \le 4$, and $x_3 \le 2$. If we used the standard formula, we would accidentally count scenarios where we place 5 X's in the top row, which is physically impossible. When constraints are tight and small, case enumeration is often the most robust and error-free path.

Q: Is there a faster way than listing all the cases? This feels tedious.

I understand the desire for a 'shortcut,' but in JEE Advanced, the most dangerous trap is trying to apply a complex formula to a simple problem. Because the total number of X's is small (6) and the number of rows is small (3), listing the cases is actually the fastest and safest method. It prevents the 'over-counting' errors that often plague more abstract methods like the Principle of Inclusion-Exclusion.

Q: How do I know if I have missed a case?

The best way to ensure you haven't missed a case is to be systematic. Start by fixing the value of $x_1$ (the smallest row) and then iterate through the possible values of $x_2$ and $x_3$. By keeping your variables in a fixed order, you create a logical flow that makes it impossible to skip a valid combination. If you find yourself stuck, always return to the constraints: $x_1 + x_2 + x_3 = 6$ and $1 \le x_i \le \text{capacity}_i$.

Question

Six X's have to be placed in the squares of figure below in such a way that each row contains at least one X. In how many different ways can this be done.

Sigma Percentile · Accepted Answer

### The Geometry of Constraints

Imagine you are standing before this grid. It is not a uniform board; it is a tiered structure consisting of three rows: $R_1$, $R_2$, and $R_3$.

The top row, $R_1$, is a narrow corridor of $2$ squares. The middle row, $R_2$, is the grand hall with $4$ squares. The bottom row, $R_3$, mirrors the top with $2$ squares.

We are tasked with placing $6$ identical X's. The rule is simple: every row must have at least one X. This is our 'existence constraint.'

Mathematically, we define $x_1, x_2, x_3$ as the number of X's in each row. Our governing equation is:

$$x_1 + x_2 + x_3 = 6$$

We must respect the physical reality of the grid, which imposes the following boundaries:

$$1 \le x_1 \le 2, \quad 1 \le x_2 \le 4, \quad 1 \le x_3 \le 2$$

### The Art of Partitioning

We need to find all integer triplets $(x_1, x_2, x_3)$ that satisfy our sum and our boundaries. Let us be methodical.

If $x_1 = 1$, then $x_2 + x_3 = 5$. Given $x_3 \le 2$, $x_3$ can be $1$ or $2$.

If $x_3 = 1$, then $x_2 = 4$. If $x_3 = 2$, then $x_2 = 3$. These give us two valid cases: $(1, 4, 1)$ and $(1, 3, 2)$.

Now, let $x_1 = 2$. Then $x_2 + x_3 = 4$. Again, $x_3$ can be $1$ or $2$.

If $x_3 = 1$, then $x_2 = 3$. If $x_3 = 2$, then $x_2 = 2$. This gives us two more cases: $(2, 3, 1)$ and $(2, 2, 2)$.

We have found our four pillars of truth: $(1, 4, 1)$, $(1, 3, 2)$, $(2, 3, 1)$, and $(2, 2, 2)$.

### The Combinatorial Engine

Now that we have our cases, we must calculate the number of ways to arrange the X's within each case. For any case $(x_1, x_2, x_3)$, the number of ways is the product of the combinations for each row:

$$	ext{Ways} = {}^{2}C_{x_1} 	imes {}^{4}C_{x_2} 	imes {}^{2}C_{x_3}$$

Let us calculate:

Case 1 $(1, 4, 1)$: ${}^{2}C_1 	imes {}^{4}C_4 	imes {}^{2}C_1 = 2 	imes 1 	imes 2 = 4$.

Case 2 $(1, 3, 2)$: ${}^{2}C_1 	imes {}^{4}C_3 	imes {}^{2}C_2 = 2 	imes 4 	imes 1 = 8$.

Case 3 $(2, 3, 1)$: ${}^{2}C_2 	imes {}^{4}C_3 	imes {}^{2}C_1 = 1 	imes 4 	imes 2 = 8$.

Case 4 $(2, 2, 2)$: ${}^{2}C_2 	imes {}^{4}C_2 	imes {}^{2}C_2 = 1 	imes 6 	imes 1 = 6$.

### The Final Summation

Since these cases are mutually exclusive, we simply add them to find the total number of arrangements.

$$	ext{Total Ways} = 4 + 8 + 8 + 6 = 26$$

There it is. **Twenty-six** distinct ways to populate this grid. You identified the constraints, partitioned the possibilities, and executed the combinations with precision.

Animated Solution for Mathematics - Permutations and Combinations: Six X's have to be placed in the squares of figure below in such a way that each row contains at least one X. In how many different ways can this be done.

Enter Numerical Value:

Visualized Solution

Visualizing the Grid & Row Capacities

Formulating the Mathematical Constraints

Exploring Valid Integer Partitions

Case 1: Distribution $(1, 4, 1)$

Case 2: Distribution $(1, 3, 2)$

Case 3: Distribution $(2, 3, 1)$

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

Calculating the Total Ways

The Sigma Insight: Combinations and Selection

The Geometry of Constraints

The Art of Partitioning

The Combinatorial Engine

The Final Summation

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