In the following question, we will count distinct integer solutions to the equation X_1 + X_2 + X_3 + X_4 = 35. How many solutions are there if all the variables must be non-negative? How many solutions are there if all the variables must be positive? How many solutions are there if X_1 greaterthanorequalto 2, X_2 greaterthanorequalto 4, X_3 greaterthanorequalto 0, and X_4 greaterthanorequalto 0? How many solutions are there to the equation X_1 + X_2 + X_3 + X_4 lessthanorequalto 35 in which all the X_i are non-negative?

An integer is zero (0), a positive integer (1, 2, 3, etc.), or a negative integer with a minus sign (-1, -2, -3, etc.). [1] A negative number is the additive reciprocal of the corresponding positive number. [2] In mathematics languages, sets of integers are often represented by a bold Z or blackboard bold . The integers form the smallest group and smallest ring containing the natural numbers. In algebraic number theory, integers are sometimes qualified as rational integers to distinguish them from the more general algebraic integers. In fact, (rational) integers are algebraic integers that are also rational numbers.

Let us take each xi = a₁+1

x₁+x₂+x₃+x₄ = 35

∴a₁+a₂+a₃+a₄ = 31 ,where each ai is non-negative.

Now using the stars and bars method the answer is -

([tex]\left \{ {{31 + 4 - 1} \atop {4 - 1}} \right.[/tex] = [tex]\left \{ {{34} \atop {3}} \right.[/tex]}

= [tex]\frac{34!}{3! * (34 - 3)!}[/tex]

= 5984

A particular solution of this equation corresponds to the placement of 3 addition signs in the 14 spaces between successive ones in a row of 15 ones.

Since, a particular solution of the equation corresponds to the placement of k−1 addition signs in the n−1 spaces between successive ones in a row of n ones, the number of solutions of the equation x₁ + x₂ + x₃ +⋯+xₐ=n in the positive integers is

[tex]\left \{ {{n - 1} \atop {k - 1}} \right.[/tex]}

For instance, placing an addition sign in the fourth, seventh, and twelfth spaces gives

1111+111+11111+111

which corresponds to the solution x₁ = 4, x₂ = 3, x₃ = 5, x₄ = 3.

Hence, the number of such solutions is the number of ways we can select which three of the 14 spaces between successive ones in a row of 15 ones will be filled with addition signs, which is

[tex]\left \{ {{14} \atop {3}} \right.[/tex]}

= [tex]\frac{14!}{3! * (14 - 3)!}[/tex]

= 364

Therefore, there are 364 positive variables and 5984 non - negative integers.

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