Determine How many integer solution there are to X1+X2+X3+x4=19 if
A) 0 ¡Ü Xi for all 1 ¡Ü i ¡Ü 4
B) 0¡Ü Xi < 8 for all 1 ¡Ü i ¡Ü 4
C) 0¡ÜX1,¡Ü5,0¡ÜX2¡Ü6,3¡ÜX3¡Ü7,3¡ÜX4¡Ü8
To determine the number of integer solutions to the equation X1 + X2 + X3 + X4 = 19 for the given conditions, we can use the concept of combinatorics and generating functions.
A) 0 ≤ Xi for all 1 ≤ i ≤ 4:
In this case, we can think of the problem as distributing 19 identical objects (represented by 19 stars, *) into 4 distinct groups (represented by the 4 variables X1, X2, X3, X4) with each group having at least 0 objects. This problem can be solved using stars and bars method.
Let's consider the variables X1, X2, X3, X4 as placeholders for the objects between two stars as shown below:
*X1* X2* X3* X4*
Now, let's consider the case of having 19 stars. We need to find the number of ways to arrange these 19 stars with the placeholders in between.
For this problem, we need to choose 3 placeholders from the 18 available spots (19 stars - 1 spot for each variable X1, X2, X3, X4) to split the stars into 4 groups. This can be expressed as choosing 3 spots from 18, which can be calculated using the formula for combinations:
C(18, 3) = 18! / (3!(18 - 3)!) = 816
Therefore, there are 816 integer solutions.
B) 0 ≤ Xi < 8 for all 1 ≤ i ≤ 4:
In this case, the problem is similar to case A, but with an additional constraint that each variable Xi should be less than 8. We can think of this as subtracting 8 from each variable Xi so that the values become non-negative.
Let's say Y1 = X1 - 8, Y2 = X2 - 8, Y3 = X3 - 8, Y4 = X4 - 8. Now we have Y1, Y2, Y3, Y4 satisfying 0 ≤ Yi < 8.
Using the same approach as in case A, we have 19 - 4(8) = 19 - 32 = -13 stars remaining. Now, we need to distribute -13 stars into 4 groups with the constraint that each group has at least 0 stars. Since we cannot have a negative number of stars, there are no solutions to this equation under the given conditions.
Therefore, there are no integer solutions.
C) 0 ≤ X1 ≤ 5, 0 ≤ X2 ≤ 6, 3 ≤ X3 ≤ 7, 3 ≤ X4 ≤ 8:
In this case, we have specific ranges for each variable Xi. We can approach this problem by considering the following:
Let Y1 = X1, Y2 = X2, Y3 = X3 - 3, Y4 = X4 - 3. Now we have Y1, Y2, Y3, Y4 satisfying 0 ≤ Y1 ≤ 5, 0 ≤ Y2 ≤ 6, 0 ≤ Y3 ≤ 4, 0 ≤ Y4 ≤ 5.
Now, let's find the number of solutions for this new equation Y1 + Y2 + Y3 + Y4 = 19 - 2(3) = 13 using the same approach as in case A.
We need to distribute 13 stars into 4 groups, which can be solved using the stars and bars method. Choosing 3 placeholders from the 12 available spots:
C(12, 3) = 12! / (3!(12 - 3)!) = 220
Therefore, there are 220 integer solutions.