Question

How many solutions are there to the equation \(x_1 + x_2 + x_3 + x_4 = 17\), where \(x_1, x_2, x_3, x_4\) are nonnegative integers?

• A. 1140
• B. 1160
• C. 1040
• D. 1200

Hint:

The "stars and bars" theorem helps in counting the number of solutions to equations of the form \(x_1 + x_2 + ... + x_n = k\) where \(x_1, x_2, ..., x_n\) are non-negative integers.

Answer 1

The Correct Option is A

1. Identify the Problem Type: We need to find the number of non-negative integer solutions to the linear equation \(x_1 + x_2 + x_3 + x_4 = 17\). This is a classic combinatorial problem that can be solved using the "stars and bars" method.
2. Stars and Bars Method Explanation:
   • Imagine the sum 17 as 17 identical items (stars): \(* * * * * * * * * * * * * * * * *\)
   • We need to divide these 17 stars into 4 distinct bins (representing the variables \(x_1, x_2, x_3, x_4\)). To separate the stars into 4 bins, we need \(4 - 1 = 3\) dividers (bars).
   • For example, the arrangement \( ** | ***** | ******* | *** \) would correspond to the solution \(x_1=2, x_2=5, x_3=7, x_4=3\).
   • The problem is equivalent to finding the number of distinct ways to arrange the 17 stars and 3 bars in a sequence.
3. Formulate as a Combination Problem:
   • We have a total of \(17 \text{ (stars)} + 3 \text{ (bars)} = 20\) positions in the sequence.
   • We need to choose the positions for the 3 bars (the remaining positions will automatically be filled by stars).
   • The number of ways to do this is given by the combination formula \(\binom{n}{k}\), where \(n\) is the total number of positions and \(k\) is the number of items to choose (in this case, the bars).
4. Apply the Formula:
   • Total positions \(n = 17 + 4 - 1 = 20\).
   • Number of bars to place \(k = 4 - 1 = 3\).
   • The number of solutions is \(\binom{20}{3}\).
   • Alternatively, we can think of choosing the positions for the 17 stars out of 20 total positions, which is \(\binom{20}{17}\). Note that \(\binom{20}{3} = \binom{20}{17}\).
5. Calculate the Combination:

\[ \binom{20}{3} = \frac{20!}{3!(20-3)!} = \frac{20!}{3!17!} \]

\[ = \frac{20 \times 19 \times 18 \times 17!}{ (3 \times 2 \times 1) \times 17!} \]

Cancel out the \(17!\):

\[ = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = \frac{20 \times 19 \times 18}{6} \]

Simplify:

\[ = 20 \times 19 \times 3 = 60 \times 19 \]

\[ = \mathbf{1140} \]

1. Compare with Options: The calculated number of solutions is 1140. This matches option (A).

The correct option is (A) 1140.

Answer 2

We are given the equation: \[ x_1 + x_2 + x_3 + x_4 = 17 \] where \( x_1, x_2, x_3, x_4 \) are non-negative integers. 

This is a classic stars and bars problem. The number of non-negative integer solutions to: \[ x_1 + x_2 + \dots + x_k = n \] is given by: \[ \binom{n + k - 1}{k - 1} \] In our case, \( n = 17 \) and \( k = 4 \). So the number of solutions is: \[ \binom{17 + 4 - 1}{4 - 1} = \binom{20}{3} = \frac{20 . 19 . 18}{3 \cdot 2 \cdot 1} = 1140\]

Correct answer: (A) 1140

Answer 3

This is a classic example of the stars and bars problem, where the total sum is 17 and we are partitioning it into 4 nonnegative integers. The formula for the number of solutions is: \[ \text{Number of solutions} = \binom{17 + 4 - 1}{4 - 1} = \binom{20}{3} \] Now, calculating: \[ \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140 \] Hence, the number of solutions is 1140.