How Many Ways Can Rs. 16 Be Divided Among 4 People?

Assume the four people are distinct, the rupees are indivisible whole units, and “none of them get” means that nobody gets zero. The problem is then to count the ways to give each person a positive integer number of rupees totaling 16.

The answer is 455. The stars-and-bars method gives this count directly; the key is to count allocations to named people, not just different sets of four amounts.

Turn the problem into an equation

Let $x_1,x_2,x_3,x_4$ be the amounts received by the four people. We need to count the positive-integer solutions of

$$x_1+x_2+x_3+x_4=16, \qquad x_i\geq 1.$$

For example, $(1,5,4,6)$ is one allocation: the first person gets Rs. 1, the second Rs. 5, and so on. The allocation $(5,1,4,6)$ is a different one because the first and second people have swapped amounts.

Reserve one rupee for everyone

Give each person Re. 1 before distributing anything else. This uses Rs. 4 and leaves Rs. 12. Define $y_i=x_i-1$ as the extra amount person $i$ receives. Now each $y_i$ may be zero, and

$$y_1+y_2+y_3+y_4=12, \qquad y_i\geq 0.$$

The problem has become a standard stars-and-bars count: distribute 12 identical rupees among four labeled people, allowing some people to receive no extra rupees.

Apply stars and bars

Represent the 12 remaining rupees by 12 stars in a row. Three bars divide the row into four groups, one for each person. For instance,

$$**\mid *****\mid \mid *****$$

represents extra amounts $(2,5,0,5)$, so the original shares are $(3,6,1,6)$.

There are 12 stars and 3 bars, for 15 positions altogether. Choose which 3 positions contain bars; the remaining positions contain stars:

$$\binom{12+4-1}{4-1}=\binom{15}{3} =\frac{15\cdot14\cdot13}{3\cdot2\cdot1}=455.$$

So there are 455 allocations under these assumptions.

If the minimum share is different

The wording can also mean that each person must receive at least some specified amount. If each person must receive at least $m$ whole rupees, first reserve $m$ rupees for each person. Provided $16\geq 4m$, the count is

$$\binom{16-4m+3}{3}.$$

For example, if each person must receive at least Rs. 2, the remaining Rs. 8 can be distributed freely, giving

$$\binom{8+3}{3}=\binom{11}{3}=165.$$

If the four recipients are not distinguished from one another, the question instead asks for partitions of 16 into four positive parts; that is a different counting problem. Likewise, if any fractional amount of a rupee is allowed, a smallest unit (such as paise) must be specified to get a finite count.

Common mistakes

• Allowing zero shares: The positive-share condition excludes allocations where someone receives nothing. The initial one-rupee reservation enforces that condition.
• Treating the people as interchangeable: If the people are distinct, assigning Rs. 2, Rs. 4, Rs. 4, and Rs. 6 to different people can yield several allocations. Do not count only unordered lists of amounts.
• Using the wrong number of dividers: Four recipients require three bars, because three dividers split a row into four groups.
• Confusing a share minimum with an exact share: “At least Rs. 2” allows a person to receive more than Rs. 2; subtract the minimum for every person before counting the remainder.