Math Problem Statement

How many solutions satisfy 𝑎 + 𝑏 + 𝑐 + 𝑑 ≤ 10 𝑎, 𝑏, 𝑐, 𝑑 ≥ 1

Solution

To solve the problem of finding the number of solutions that satisfy:

a+b+c+d10witha,b,c,d1,a + b + c + d \leq 10 \quad \text{with} \quad a, b, c, d \geq 1,

we'll transform the inequality and then apply the combinatorial "stars and bars" method.

Step 1: Substitute Variables

Since each variable a,b,c,a, b, c, and dd must be at least 1, we can let: a=a1,b=b1,c=c1,d=d1,a' = a - 1, \quad b' = b - 1, \quad c' = c - 1, \quad d' = d - 1, where a,b,c,a', b', c', and d0d' \geq 0.

Now the inequality becomes: (a+1)+(b+1)+(c+1)+(d+1)10.(a' + 1) + (b' + 1) + (c' + 1) + (d' + 1) \leq 10.

Simplifying, we get: a+b+c+d6.a' + b' + c' + d' \leq 6.

Step 2: Convert to an Equality with a New Variable

Let e=6(a+b+c+d)e = 6 - (a' + b' + c' + d'), where e0e \geq 0. Then, the equation becomes: a+b+c+d+e=6.a' + b' + c' + d' + e = 6.

Now, we need to find the number of non-negative integer solutions to this equation, which is a classic combinatorial problem.

Step 3: Apply the Stars and Bars Theorem

The number of solutions to the equation x1+x2+x3+x4+x5=nx_1 + x_2 + x_3 + x_4 + x_5 = n in non-negative integers is given by: (n+k1k1),\binom{n + k - 1}{k - 1}, where n=6n = 6 and k=5k = 5 (since we have a,b,c,d,a', b', c', d', and ee).

Thus, the number of solutions is: (6+5151)=(104).\binom{6 + 5 - 1}{5 - 1} = \binom{10}{4}.

Step 4: Calculate (104)\binom{10}{4}

(104)=10!4!(104)!=10×9×8×74×3×2×1=210.\binom{10}{4} = \frac{10!}{4!(10 - 4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210.

Conclusion

The number of solutions that satisfy a+b+c+d10a + b + c + d \leq 10 with a,b,c,d1a, b, c, d \geq 1 is:

210.\boxed{210}.

Would you like more details on any part of this solution?


Here are five related questions that expand on this concept:

  1. How would the solution change if the inequality were a+b+c+d=10a + b + c + d = 10?
  2. How does the "stars and bars" method generalize to more variables?
  3. What would happen if we imposed a stricter upper bound on each variable individually?
  4. How can we apply similar methods to find integer solutions within a range for each variable?
  5. How would the solution change if we allowed one of the variables to be zero?

Tip: The stars and bars method is a powerful technique for counting integer solutions to equations, especially useful when dealing with combinatorial inequalities.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Inequalities
Stars and Bars Method

Formulas

Substitution for non-negative variables: x' = x - 1
Stars and Bars formula: C(n + k - 1, k - 1)

Theorems

Stars and Bars Theorem

Suitable Grade Level

Grades 9-12