This problem can be solved using the stars and bars method in combinatorics.

The stars and bars method is a way to distribute n identical items into k bins. In this case, we are distributing 38 identical items (the total sum) into 3 bins (the variables x1, x2, x3).

The formula for stars and bars is (n+k-1 choose k-1), where "choose" represents a combination.

So, we have (38+3-1 choose 3-1) = (40 choose 2).

The combination formula is n! / [k!(n-k)!], where "!" represents factorial.

So, (40 choose 2) = 40! / [2!(40-2)!] = 40*39 / (2*1) = 780.

Therefore, there are 780 possible ordered triples (x1, x2, x3) such that x1, x2, x3 are non-negative integers and x1+x2+x3=38.

Question

This problem can be solved using the stars and bars method in combinatorics.

The stars and bars method is a way to distribute n identical items into k bins. In this case, we are distributing 38 identical items (the total sum) into 3 bins (the variables x1, x2, x3).

The formula for stars and bars is (n+k-1 choose k-1), where "choose" represents a combination.

So, we have (38+3-1 choose 3-1) = (40 choose 2).

The combination formula is n! / [k!(n-k)!], where "!" represents factorial.

So, (40 choose 2) = 40! / [2!(40-2)!] = 40*39 / (2*1) = 780.

Therefore, there are 780 possible ordered triples (x1, x2, x3) such that x1, x2, x3 are non-negative integers and x1+x2+x3=38.

Knowee AI · Accepted Answer

This problem can be solved using the stars and bars method in combinatorics.

The stars and bars method is a way to distribute n identical items into k bins. In this case, we are distributing 38 identical items (the total sum) into 3 bins (the variables x1, x2, x3).

The formula for stars and bars is (n+k-1 choose k-1), where "choose" represents a combination.

So, we have (38+3-1 choose 3-1) = (40 choose 2).

The combination formula is n! / [k!(n-k)!], where "!" represents factorial.

So, (40 choose 2) = 40! / [2!(40-2)!] = 40*39 / (2*1) = 780.

Therefore, there are 780 possible ordered triples (x1, x2, x3) such that x1, x2, x3 are non-negative integers and x1+x2+x3=38.

How many possible ordered triples (𝑥1,𝑥2,𝑥3)(𝑥1,𝑥2,𝑥3) are there such that 𝑥1,𝑥2,𝑥3𝑥1,𝑥2,𝑥3 are non-negative integers and 𝑥1+𝑥2+𝑥3=38𝑥1+𝑥2+𝑥3=38?

Question

How many possible ordered triples $(x_1,x_2,x_3)$ are there such that $x_1,x_2,x_3$ are non-negative integers and $x_1+x_2+x_3=38$ ?

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