Counting Quadruples of Integers: A Combinatorial Problem

[YAHA]

This should be a simple combinatorial problem. Suppose I have a number n which is a positive integer. Suppose, that there are four numbers a,b,c,d such that 0<=a<=b<=c<=d<=n.
The question is how many quadruples of the form (a,b,c,d) can be formed out such arrangement?

I realize that this is a homework-like question, but I am really interested in seeing which principles of combinatoris would apply here.

 

[chiro]

Hey YAHA.

Recall that if you get a number x on a particular draw, then the number of possibilities in the next draw will be n-x.

The rest is to apply the multiplication rule on these to get the number of possibilities for one particular set of observations.

Then sum over all possibilities and this will give you the total number of combinations.

So as an example, you look at the (n+1)*(n+1-x)*(n+1-y)*(n+1-z) for getting (*,x,y,z).

Now sum all of these over all consistent values of x, y, and z and that will give you the number of combinations.

 

[awkward]

The number of quadruples is the number of multisets of 4 numbers drawn from the integers 0, 1, 2, ..., n (n+1 integers), hence

$$\binom{n+1+4-1}{4} = \binom{n+4}{4}$$

 

[haruspex]

The number of quadruples is the number of multisets of 4 numbers drawn from the integers 0, 1, 2, ..., n (n+1 integers), hence

$$\binom{n+1+4-1}{4} = \binom{n+4}{4}$$

Quite so, but I suspect that will need more detailed explanation. Here's my version, yours may be simpler.
We can map the problem into one where a< b < c < d, merely by adding 0, 1, 2, 3 respectively. Since we added 3 to d, we must also add 3 to n. So now it's a matter of the number of ways of choosing 4 distinct numbers from 0,..,n+3.

 

[blue_raver22]

I understand the importance of combinatorics in solving problems related to counting and arranging objects. In this case, the problem can be solved using the principles of combinations and permutations.

Firstly, we can see that the values of a, b, c, d are restricted to be between 0 and n, with a being the smallest and d being the largest. This suggests that we are dealing with combinations, as the order of the numbers does not matter.

To find the number of possible combinations, we can use the formula for combinations, which is nCr = n! / (r!(n-r)!). In this case, we have 4 numbers to choose from (a, b, c, d) and we need to choose 4 of them, so n = 4 and r = 4. Thus, the total number of combinations would be 4C4 = 4! / (4!(4-4)!) = 1.

However, this only gives us the total number of combinations without considering the restrictions on the values of a, b, c, d. To account for these restrictions, we need to use the principle of inclusion-exclusion.

Firstly, we need to consider the restriction that 0 <= a <= b <= c <= d <= n. This means that the values of a, b, c, d can be chosen from a range of n+1 numbers (0, 1, 2, ..., n). Therefore, the total number of combinations without any restrictions would be (n+1)C4.

However, this includes combinations where a, b, c, d are not in increasing order. To account for this, we need to subtract the number of combinations where the values are not in increasing order. This can be calculated by considering the number of ways to arrange 4 numbers in decreasing order, which is 4!. Therefore, the total number of combinations with this restriction would be (n+1)C4 - 4!.

Similarly, we need to consider the other restrictions on the values of a, b, c, d. This would include the restriction that a, b, c, d cannot be greater than n. This can be calculated by considering the number of combinations where at least one of the values is greater than n, which would be (n+1)C4 - (n+1).

Finally, we need to add back the number of combinations