ML Interview Q Series: Grid Placement Probability: Calculating Unique Row/Column Arrangements Using Combinatorics
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Question: Five dots are placed at random on an 8×8 grid such that no cell contains more than one dot. Specify an appropriate sample space and determine the probability that no row or column contains more than one dot.
Short Compact solution
One way to view the sample space is to consider all ordered choices of 5 distinct cells in the 8×8 grid. The total number of outcomes is 64×63×62×61×60. Among these, the number of favorable cases where no row or column contains more than one dot is 64×49×36×25×16. Hence, the probability is (64×49×36×25×16) / (64×63×62×61×60), which is approximately 0.0494.
An alternative approach uses an unordered sample space of all ways to choose 5 different cells from 64 cells, so the total number of outcomes is (64 choose 5). Then the number of ways to choose 5 cells with at most one dot in any row or column can be computed by first picking which 5 rows to use, then choosing one cell in each of those rows such that all columns are distinct. That count is (8 choose 5)×8×7×6×5×4, giving the same probability of about 0.0494.
Comprehensive Explanation
When placing 5 dots on an 8×8 grid with the constraint that no cell holds more than one dot, we can tackle the problem through two primary views: an ordered sample space or an unordered sample space. Both perspectives, if handled correctly, lead to the same probability.
Ordered sample space perspective
In an ordered framework, we imagine choosing the first dot from any of the 64 cells, then the second from any remaining 63 cells, the third from any remaining 62 cells, and so forth. This counts every distinct sequence of 5 cell placements as a separate outcome.
Total number of outcomes in the ordered perspective: 64×63×62×61×60.
Under the condition that no row or column contains more than one dot, each new dot has restricted placement options. Specifically:
For the first dot, there are 64 possibilities (no restrictions yet).
For the second dot, we can no longer use the row or column of the first, so that leaves 7 rows and 7 columns that are still free. Thus the second dot has 49 possibilities.
For the third dot, we lose another row and another column, leaving 6×6 = 36 possibilities.
For the fourth dot, we have 5×5 = 25 possibilities.
For the fifth dot, we have 4×4 = 16 possibilities.
So the number of ways (outcomes) that satisfy the row-column restriction is 64×49×36×25×16. The probability is the ratio of these favorable outcomes to the total outcomes:
which yields approximately 0.0494.
Unordered sample space perspective
It is often more intuitive to treat the problem in an unordered way. Instead of considering sequences of placements, we only care about which set of cells (5 unique cells) is chosen, irrespective of the order in which they were selected.
Total number of ways to pick 5 different cells (unordered) out of 64: (64 choose 5).
Next, to count how many such sets satisfy the “no row or column repetition” constraint, note that each selected dot must occupy a different row and a different column. The strategy is:
Choose 5 distinct rows from the 8 available rows. There are (8 choose 5) ways to do so.
In each of these chosen rows, we must choose exactly one cell, with no two chosen cells falling in the same column. This is equivalent to assigning distinct columns to each chosen row. The number of ways to assign 5 distinct columns to these 5 rows is 8×7×6×5×4 if the order in which we pair rows to columns matters. That factor 8×7×6×5×4 is the number of ways to assign columns to the first chosen row, then the second, etc., each time avoiding a column already used.
Hence, the favorable configurations are counted by (8 choose 5) × (8×7×6×5×4). Dividing this count by (64 choose 5) produces the same probability of approximately 0.0494:
Why both approaches agree
Both methods count the same “physical” configurations of dots on the grid, but the first method distinguishes them by the order in which the dots are placed, while the second sees them as combinations. As long as one is consistent in counting the total sample space and the favorable outcomes under the same perspective (ordered vs. ordered, or unordered vs. unordered), they yield the same result.
Potential Follow-up Questions
What if the grid were n×n instead of 8×8?
If the grid is n×n and we place k dots such that no two lie in the same row or column, the same logic extends. Ordered sample space (if no cell can be used twice) is n×(n−1)×(n−2)×... up to k terms. The number of favorable ways (ordered) is n×(n−1)×(n−2)×... up to k terms, but factoring in the row–column restriction changes the count as soon as one dot is placed. We get:
Total outcomes (ordered): n×(n−1)×...×(n−k+1).
Favorable outcomes (ordered), if no two dots share row or column: n×(n−1)×...×(n−k+1) for placing distinct dots in distinct rows and columns. More precisely, each new dot “uses up” one row and one column, so it is n×(n−1)×(n−2)×...×(n−k+1) as well, but in a systematically restricted manner (similar to permutation of n objects taken k at a time).
Probability is the ratio of those two, which in the distinct-row/column scenario effectively becomes the fraction of permutations over permutations, giving the same result.
If viewed in an unordered manner, we use the combination approach: total ways to pick k distinct cells is (n² choose k). Favorable ways to ensure no row or column overlap is (n choose k) times the number of ways to assign these k rows to k distinct columns, which is P(n, k) = n×(n−1)×...×(n−k+1). Dividing by (n² choose k) yields the probability.
How can we implement a brute-force check in Python?
We can do a short enumeration in Python if n = 8 and k = 5 is not too large, though it is still somewhat expensive. A direct approach would be:
import math
def brute_force_probability(n=8, k=5):
from itertools import combinations
cells = [(r, c) for r in range(n) for c in range(n)]
total_ways = 0
valid_ways = 0
for subset in combinations(cells, k):
total_ways += 1
rows_used = set()
cols_used = set()
valid = True
for (r, c) in subset:
if r in rows_used or c in cols_used:
valid = False
break
rows_used.add(r)
cols_used.add(c)
if valid:
valid_ways += 1
return valid_ways / total_ways
print(brute_force_probability()) # Should be roughly 0.0494
This uses combinations from Python’s standard library, iterates over all ways to place 5 dots in an 8×8 grid, and checks if no row or column is repeated. It then computes the fraction that satisfies our constraint.
How does this relate to placing rooks on a chessboard?
Placing dots with no two in the same row or column is analogous to placing non-attacking rooks on a chessboard. In combinatorics, the number of ways to place k non-attacking rooks on an n×n board is (n choose k) multiplied by k! if we consider permutations of columns for a chosen set of rows. This is exactly the kind of counting we are doing.
Are there any real-world data-related scenarios where this combinatorial logic is relevant?
Yes. The notion of ensuring distinct row and column selections arises in multiple scheduling and assignment optimization tasks. For example, if each dot corresponds to scheduling a person in a time slot, then “no row or column overlap” might represent ensuring that a person does not appear twice in a schedule or that a room/time combination is uniquely assigned. Combinatorial counting and permutations are fundamental tools in these contexts.
How might mistakes occur in counting?
Common pitfalls:
Mixing the ordered approach for counting favorable cases with the unordered approach for counting the total sample space. This leads to mismatched denominators and numerators.
Forgetting that each new dot restricts both a row and a column and accidentally counting fewer or more possibilities than exist.
Confusing permutations with combinations when distributing k indistinguishable or distinguishable items.
Being consistent in how you count total outcomes versus favorable outcomes is crucial to avoid incorrect results.
Below are additional follow-up questions
What if we changed the condition so that at most two dots can occupy the same row or column? How would that change the counting approach?
If we allow up to two dots in any given row or column (instead of only one), the counting framework needs to be modified to capture that relaxed constraint.
Detailed Explanation
In this modified scenario, a single row or column might have zero, one, or two dots, but it must not exceed two. We can no longer rely on simple permutations for counting valid placements. Instead, an approach might involve:
Counting all possible ways to place k dots (where k=5 in the original problem) with no more than two dots in each row or column.
Dividing by the total number of ways to place k dots in any distinct cells.
A systematic method is:
Case-by-case enumeration: We can split into cases based on how many rows have two dots and how many rows have one dot (and similarly for columns). For instance, for 5 dots, possible row distributions include:
5 rows each having 1 dot (with zero rows having 2 dots),
1 row having 2 dots and 3 rows having 1 dot,
2 rows having 2 dots and 1 row having 1 dot, etc.
For each case, we would check how columns accommodate those distributions. The complexity can grow quickly, but combinatorial arguments or generating functions can be employed.
A pitfall here is oversimplification. One might try to treat it as “choose a row distribution, choose column distribution,” but double counting or undercounting can occur when you assume independence between row and column placements. Properly enumerating these configurations or using a structured combinatorial approach (e.g., the principle of inclusion-exclusion) can be vital. In real-world problems, miscounting due to relaxed constraints is common because the neat permutations used for the “at most one dot” scenario no longer apply directly.
What if the dots have additional attributes (for instance, each dot has a color) and we care about arrangements with no row/column collisions for dots of the same color?
When dots are no longer identical, we must factor their attributes or labels into the probability calculation. For example, if each dot has a distinct color, the order or arrangement in which those colors are placed on the board becomes relevant.
Detailed Explanation
Sample space: If each of the 5 dots is distinguishable (say 5 different colors), the total number of ways to place these dots on an 8×8 grid (with at most one dot per cell) might still be 64×63×62×61×60 if we think in an ordered way. However, the color arrangement matters. If we are looking at unordered placements of cells but still want to keep color labels, then the count of total placements is (64 choose 5)×5! since we are choosing 5 cells for the 5 dots and then permuting the colored dots among those chosen cells.
Favorable outcomes: We want no row or column containing more than one dot of the same color, but potentially it could contain different colors. This situation is less restrictive than the typical “only one dot total per row/column.” You might allow, for instance, the same row to have a red and a blue dot, but not two red dots.
Pitfalls:
Mixing up conditions like “no row can contain two dots of the same color” vs. “no row can contain more than one dot total.”
Failing to account for permutations of dot colors vs. cell choices.
For multi-attribute dots, carefully structuring the sample space (ordered vs. unordered) and clarifying whether each dot is truly unique or only partially unique by color is essential. A subtle but common mistake is to treat dots as identical when they are not, or vice versa.
How does the probability scale as the number of placed dots grows larger compared to the size of the grid?
When more dots are placed, the probability of having no row or column collisions typically decreases. For an n×n grid, once you exceed n dots, it is impossible to have a collision-free arrangement. Even for k < n, as k gets closer to n, the probability can become extremely small.
Detailed Explanation
Low k: If k is small relative to n (for instance, k << n), collisions are unlikely. Combinatorial formulas suggest a sizable fraction of placements remain collision-free.
Moderate k: As k increases, the chance of collision-free placements drops. The typical approach is to compare the number of permutations allowed by row and column constraints to the total ways of picking cells.
High k: Approaching k = n, we are essentially looking at permutations or partial permutations (like placing n rooks on an n×n board without collision). The number of favorable ways is n! if k = n (ordered), and the total ways to place n distinct dots is (n² choose n) for the unordered space. The ratio n! / (n² choose n) becomes quite small for large n.
A common pitfall is to neglect the combinatorial explosion in the total number of ways (n² choose k) while focusing only on intuitive guesses about collisions. In practical data scenarios with large n and non-trivial k, direct enumeration becomes infeasible, and approximate or asymptotic methods might be necessary.
How might we use inclusion-exclusion to verify or derive the probability?
Inclusion-exclusion is a fundamental combinatorial tool that can be used to count the number of ways to arrange items subject to constraints (such as “no two dots in the same row or column”).
Detailed Explanation
Setup: We define events like E_r = “row r has at least two dots” and E_c = “column c has at least two dots.” Then we want the probability of none of these events occurring.
Inclusion-exclusion formula: We calculate the total number of ways to place the dots minus the sum of placements violating at least one row or column constraint, plus the sum of placements violating at least two row/column constraints, minus the sum of those violating three constraints, and so on.
Pitfalls:
The number of terms in the formula can be large: we have 8 row events plus 8 column events, for a total of 16. We need to consider intersections of these events, which can explode combinatorially.
Overcounting or undercounting can happen if we fail to carefully handle intersections like “row r1 has two dots” and “column c1 has two dots.”
While the direct permutation or combination approach is more straightforward for the “no collisions” condition, using inclusion-exclusion is often a valuable cross-check in advanced combinatorial proofs or in more complicated constraints.
Can we generalize to three-dimensional or higher-dimensional grids?
Yes, we can extend the concept to placing points in a three-dimensional grid (e.g., an n×n×n lattice) or higher dimensions, with constraints like “no two points share the same plane” or “no two points share the same line in any dimension.”
Detailed Explanation
3D lattice: Suppose you have an n×n×n grid of cells and you place k distinct points. A possible constraint might be that no two points can share the same x-coordinate, y-coordinate, or z-coordinate. The counting method parallels our 2D scenario but each point excludes a row, column, and “vertical” slice in 3D. Formally, you are limiting any dimension’s coordinate from repeating.
Pitfalls: The combinatorial expressions escalate in complexity because each point restricts a full hyperplane in higher-dimensional space. For 2D, each dot eliminates one row and one column. In 3D, each point eliminates a row, column, and “depth,” so effectively 3 constraints.
Real-world analogy: This might appear in resource allocation problems with multiple constraints or in certain scheduling tasks where you have multiple “dimensions” like time slot, room, teacher, etc., and you want no collisions across any dimension. Managing collisions in higher dimensions often requires careful combinatorial or algorithmic solutions (e.g., bipartite or multipartite matching in more advanced formulations).
How do rounding errors or floating-point precision issues come into play when computing these probabilities in practice?
When probabilities become very small or involve large factorials, floating-point arithmetic in a computer can cause inaccuracies or underflow/overflow errors.
Detailed Explanation
Factorials and binomial coefficients can get extremely large even for moderate n. Directly computing (64 choose 5) in Python is safe because 64 choose 5 is not astronomically large, but for bigger n, the intermediate calculations can be huge.
Pitfalls:
If you convert each term to floating-point prematurely, you may lose significant precision, especially if the ratio of two large numbers is the ultimate quantity of interest.
Summations of large terms in inclusion-exclusion can also cause issues where intermediate terms exceed floating-point ranges.
Possible solutions:
Use exact arithmetic libraries or symbolic computation for combinatorial expressions.
Simplify the ratio analytically before computing it numerically. For instance, we can reduce factorials or products that appear in both the numerator and denominator to keep numbers manageable.
How does this probability problem relate to random graph theory or adjacency matrix constraints?
Placing dots on an 8×8 grid with no shared rows or columns can be viewed through the lens of adjacency matrices in bipartite graphs or permutation matrices in discrete mathematics.
Detailed Explanation
Permutation matrix: A matrix with exactly one “1” per row and column (and 0s elsewhere) is known as a permutation matrix. In our scenario with 5 dots, we effectively create a matrix with 5 ones and the rest zeros, but no row or column can have more than one “1.” So we have a partial permutation matrix.
Connection to bipartite matching: Each row can be seen as one set of vertices, each column as the other set, and a dot indicates an edge. The requirement that no row or column has more than one dot is equivalent to having a matching of size 5 in a bipartite graph.
Pitfalls:
Failing to distinguish between a complete matching (one that covers all vertices on one side) and partial matching.
Confusion about whether the row set and the column set each must match exactly once. For a partial matching of size k < n, not all vertices on the bipartite graph are used.
Real-world example: Resource allocation among two groups (tasks and machines, for instance) can be seen as bipartite matching. Probability questions might arise if you randomly generate edges in a bipartite graph and wonder about the probability of a certain-size matching.
What if the random process of placing dots is sequential, but each placement is non-uniform (e.g., different cells have different probabilities)?
So far, we typically assume each distinct arrangement is equally likely. In some scenarios, each cell might have a distinct probability of being chosen, meaning the distribution is non-uniform.
Detailed Explanation
Weighting each cell: Suppose each of the 64 cells has some probability p_ij where i and j range over rows and columns, subject to the constraint that the probabilities sum to 1 if you only place one dot, or that each new dot is placed with probabilities adjusted after each selection.
Collision constraints: Ensuring that no row or column is used more than once in a non-uniform model means each subsequent placement modifies the probability distribution for remaining cells.
Pitfalls:
Markov chain or dependency: The probability for the second dot’s location depends on where the first dot went, making the scenario more complex than simple combinatorial counting. We might need dynamic programming or specialized sampling.
Practical data scenarios: You might have prior probabilities of usage for certain cells (e.g., certain row-column pairings are more likely). Real-world scheduling or sensor placement problems often have heterogeneous constraints that turn the uniform assumption invalid.
Exact computation: A direct formula for the probability of no collisions might be unavailable in closed form. Monte Carlo simulation or sophisticated enumerations might be required.
Is there a direct connection to the concept of derangements, and could it be misapplied here?
A derangement is a permutation of n elements where none of the elements appear in their “natural” position. This might seem related to “no row or column collisions,” but it is actually quite different.
Detailed Explanation
Derangement: Typically, if we label rows and columns from 1 to n, a derangement would place the dot for row i in some column j where j ≠ i, for each i.
Collision-free arrangement: The condition we are dealing with is that no row or column is used more than once overall. That does not necessarily mean row i cannot use column i; it just means row i can’t share that column with row j.
Pitfalls:
One might incorrectly think “We need to avoid row=column for each dot,” which would count only placements where the diagonal is disallowed. That is not the correct interpretation of “no row or column collisions.”
Mixing up these concepts can lead to significantly wrong answers in probability or combinatorial arguments.
When presenting solutions in interviews, clarifying the difference between “no two dots share the same row or column” and “no dot is placed in its ‘own’ position (derangement)” is important to avoid confusion.
How would you modify the brute-force code to handle partial constraints, such as restricting collisions only in rows but allowing collisions in columns?
If we only forbid collisions in rows but allow multiple dots in the same column, the validation check changes.
Detailed Explanation
In the Python enumeration approach, for each combination of cells, we only verify that each row is used at most once. That means:
rows_used = set()
valid = True
for (r, c) in subset:
if r in rows_used:
valid = False
break
rows_used.add(r)
We do not track columns_used because collisions in columns are now allowed. The rest of the code can remain the same. The resulting probability will be higher than before because relaxing the column constraint increases the number of valid placements.
A common oversight: People might accidentally keep checking columns when they only meant to remove the row restriction. Ensuring the condition precisely matches the new problem statement is crucial, or else the count will be incorrect.