Distribution questions sit right next to selection in the CAT syllabus, and they trip up more students because the setup changes the formula completely. Instead of picking a group from a pool, you are now spreading items across groups, and whether those items are identical, whether a group can stay empty, and whether the groups themselves are distinct all change which formula applies. Get those three questions wrong and the rest of the working is wasted. This sheet lays out every standard case, from the core stars and bars result, through integer equations with lower and upper bounds, to mixed item types and the identical-groups case that has no clean formula at all. Every box carries one worked example with real numbers, so you drill the setup and not just the formula.
1Identical Items, Empty Groups Allowed
Splitting identical items into groups where a group can stay empty.
Ways = (n + r − 1)C(r − 1)
Example. 10 identical sweets among 3 kids, any kid can get zero, gives (10+3−1)C(3−1) = 12C2 = 66 ways.
CAT Hack. This is the stars and bars formula. n stars, r − 1 bars split them into r groups.
2Identical Items, Every Group Non-Empty
Splitting identical items so every group gets at least one.
Ways = (n − 1)C(r − 1)
Example. 10 identical sweets among 3 kids, each kid gets at least one, gives (10−1)C(3−1) = 9C2 = 36 ways.
CAT Hack. Hand out one item to each group first, then distribute the rest with empty groups allowed.
3Distinct Items into Distinct Groups
Each distinct item independently picks one of the distinct groups.
Ways = rn, for n items and r groups
Example. 5 distinct letters into 3 distinct boxes, each letter choosing a box on its own, gives 35 = 243 ways.
Common Mistake. Do not use nCr here. Each of the n items makes its own independent choice of group.
4Distinct Items, Non-Empty Distinct Groups
Distinct items into distinct groups with no group left empty.
Ways = rn − rC1(r−1)n + rC2(r−2)n − …
Example. 3 distinct gifts into 2 distinct kids, neither empty, gives 23 − 2C1(1)3 = 8 − 2 = 6 ways.
CAT Insight. This inclusion-exclusion pattern is rarely needed beyond 2 or 3 groups on CAT, so keep it just for small cases.
5Non-Negative Integer Solutions
Counting whole-number solutions, zero allowed, to an equation that sums to n.
x₁+x₂+…+xₛ = n, xⁱ ≥ 0, has (n+r−1)C(r−1) solutions
Example. x + y + z = 8 with each variable 0 or more has (8+3−1)C(3−1) = 10C2 = 45 solutions.
CAT Favourite. This equation form is exactly the same count as distributing n identical items into r groups.
6Positive Integer Solutions
Counting whole-number solutions where every variable is at least 1.
x₁+x₂+…+xₛ = n, xⁱ ≥ 1, has (n−1)C(r−1) solutions
Example. x + y + z = 8 with each variable at least 1 has (8−1)C(3−1) = 7C2 = 21 solutions.
CAT Hack. Give each variable 1 upfront, so x+y+z=8 becomes x′+y′+z′=5 with zero allowed.
7Solutions with a Lower Bound
An equation where one or more variables must be at least some value k.
x ≥ k → substitute x′ = x − k, then solve as usual
Example. x + y + z = 10 with x ≥ 2 becomes x′+y+z = 8 with all ≥ 0, so (8+3−1)C(3−1) = 10C2 = 45.
CAT Hack. Shift out the minimum first with a substitution, then the equation returns to the standard non-negative case.
8Solutions with an Upper Bound
An equation where a variable cannot exceed some maximum value.
Count all solutions, then subtract the ones that break the cap
Example. x+y+z=10, x ≤ 4, all ≥ 0, has total 12C2 = 66, minus the x ≥ 5 case (set x′=x−5) 8C2 = 28, giving 66−28 = 38.
CAT Insight. An upper bound almost always means total minus the complement, not a fresh direct count.
9Identical Items into Identical Groups
Both the items and the groups are indistinguishable from each other.
Count the integer partitions of n into at most r parts
Example. 8 identical candies into 3 identical bags, some bags may be empty, matches the partitions of 8 into at most 3 parts, giving 10 ways.
Common Mistake. No clean formula exists here. List the partitions directly, since n and r stay small on CAT.
10Distributing to People vs Boxes
Read whether the recipients are distinct people or identical containers.
People are always distinct, containers may or may not be
Example. ‘Distribute sweets among 3 children’ treats the children as distinct, while ‘pack sweets into 3 identical bags’ does not.
Common Mistake. The word choice in the question, children versus bags, decides which formula applies. Read it twice.
11Distributing Objects of Different Kinds
Splitting several separate collections of identical objects together.
Multiply the count for each kind's own distribution
Example. 5 identical pens and 3 identical pencils among 2 kids, each kid can get zero, gives (5+1)C1 × (3+1)C1 = 6 × 4 = 24 ways.
CAT Hack. Solve each kind of item as its own stars-and-bars problem, then multiply the two counts together.
12Distribution as a Coefficient
Reframing a distribution count as a coefficient in an expansion.
Coefficient of xn in (1+x+x2+…)r
Example. Distributing 6 identical items into 3 groups with no cap matches the coefficient of x6 in (1+x+x2+…)3, which equals 28.
CAT Insight. This generating-function view is background theory. On CAT itself, stars and bars is almost always faster.
13Choosing the Right Distribution Case
Matching the wording of a question to one of the standard setups.
Identical or distinct items, empty allowed or not, distinct or identical groups
Example. ‘At least one book to each of 4 distinct friends’ with 10 identical books signals non-empty identical items into distinct groups, so (10−1)C(4−1) = 9C3 = 84.
CAT Favourite. Underline three things before you start. Are the items identical, can a group be empty, and are the groups distinct.