Discrete Mathematics. How many solutions are there for the following equations and inequalities? All variables areintegers.Type in your answer for each part as an integer with NO COMMAS, NO SPACES, and NOLABELS.X1 + X2 + X3 + X4 + 25 = 35, such that x₂ > 0x1 + x2 + x3 ≤ 35, such that æ¡ > ix1 + x2 + x3 + x4 < 55, such that £₁ = 6, x2 > 6, x3 ≥ 0, 0 ≤ x4 < 6

Given Information:The inequations and equations are given as follows:a) .b) .c) .To find:The number of the integer solutions to each equation or inequality provided.Concept used:Fundamental counting principle:If one operation can be performed in ways and there are ways of performing a second operation, then the number of ways of performing the two operations together is .Stars and bars problem:For any pair of positive integers and , the number of of non-negative integers whose sum is is equal to the number of sets of cardinality taken from a set of size or equivalently .Combination Formula:The number of ways to choose objects from objects is calculated as .Consider the equation that .The number of integer solutions to the equation can be found using the stars-bars problem.It is required 5 non-zero integer whose sum is equal to 35.So, .Thus, the number of non-negative integer solutions to the given equation is 82251.Consider the inequation that .The number of integer solutions to the inequation can be found using the fundamental counting principle.Sum of the 4 integers are less than or equal to 35 such that .For the variable , the possibility of any integer from to . As sum is not greater than 35 with minimum entry is and other variables has at least , the inequality becomes . i.e., There are 30 ways to choose the solution to the variable .For the variable , the possibility of any integer from to .As sum is not greater than 35 with minimum entry is and other variables has at least , the inequality becomes i.e., There are 30 ways to choose the solution to the variable .For the variable , the possibility of any integer from to As sum is not greater than 35 with minimum entry is and other variable has at least , the inequality becomes i.e., There are 30 ways to choose the solution to the variable .The total number of ways to get the solution to the inequation is the number of ways of performing the all the three choices together.Thus, the number of integer solutions to the given inequation is .Consider the inequation .The number of integer solutions to the inequation can be found using the fundamental counting principle.For the variable , the possibility of the solution is only 6 as per the given condition.i.e., There are only one way to choose the solution to the variable .Now, the inequation can be re-written as follows:Now, the problem is converted to the number of integer solutions to the inequation such that .Note: As sum is less than 49 and each variable is an integer, the sum is also said as less than or equal to 48.For the variable , the possibility of any integer from 7 to 41.As sum is not greater than 48 with minimum entry is and the other variable has zero entry, the variable shouldn't be greater than 48.i.e., There are 41 ways to choose the solution to the variable .For the variable , the possibility of any integer from 0 to 41.As sum is not greater than 48 with minimum entry is and the variable has at least , the variable shouldn't be greater than .i.e., There are 41 ways to choose the solution to the variable .For the variable , the possibility of any integer from 0 to 5.As the variable ranges as , the choice of integer solutions are as .i.e., There are 6 ways to choose the solution to the variable .The total number of ways to get the solution to the inequation is the number of ways of performing the all the three choices together.Thus, the number of integer solutions to the given inequation is .a) The number of non-negative integer solutions to the given equation is .b) The number of integer solutions to the given inequation is .c) The number of integer solutions to the given inequation is .