This article was shared by 2014 MMC National Finalist Daniel James Molina. I am also fascinated with how this operator works. It is also my first time to hear that there is an H operator in combinatorics.

*“Seriously, when I read the article about Kuya Micah’s solution, (stars and bars combination thing), It really blew my mind. I can’t understand stars and bars combination, because I have another kind of solution similar to him, thanks to KUMON.”* – Daniel

We all know the Combinatorics symbols, C (non-repeated combination), P (non-repeated permutation), and the large π symbol (repeated permutation).

And know, I will share my knowledge of another symbol, which is H (repeated combination).

The identity goes like this: _{N}H_{R } = _{N+R-1}C_{R}

nHr_{ }means “ combination of n objects taken r at a time such that any of the n objects can be repeated”

**Application of the function:**

**Sample Problem 1:**

Find the number of non-negative integers that satisfy

**Solution:**

_{3}H_{15}=_{3+15-1}C_{15}=136

**Sample Problem 2:**

Find the number of positive integers that satisfy

**Solution:**

Since ,

we can find the answer by deducing it by finding the number of sets of non-negative integers that satisfy

, (A,B,C are non-negative integers)

4H11 = 14C11 = 728 sets.

What we did is changing to ,since x,y,z and w have minimal value of 1 each. At least we can change it to variables that have a minimal value of 0 for easier calculations.

[The “derivation”]

**Sample Problem 3:**

There are 30 voters and 10 candidates in an election. In how many ways can the candidates receive their votes?

**Solution:**

Let *a,b,c,d,e,f,g,h,i,j* be the candidates.

since all candidates can be given no vote,

*a,b,c,d,e,f,g,h,i*, and *j* are non-negative integers.

The number of ways is 10H30 = 39C30 = 211915132

**Sample Problem 4:**

How many ways can you choose from 5 different fruits taken 7 at a time?

**Solution:**

If you will choose 7 from the 5 fruits without repetition, you will only have 5 fruits chosen, so this problem is a repeated combination problem.

5H7 = 11C7 = 330 ways

**Sample Problem 5:**

How many ways can you group 10 people into 4 groups such that each group consist of at LEAST 2 people.

**Solution:**

Let A, B, C and D are the groups.

and

so from A + B + C + D = 10,

It will be,

4H2 = 5C2 = 10 ways.