Math  /  Algebra

QuestionGiven (a,b,c,d)(a, b, c, d) is a set of integers and all of them are greater than 6 . Find the number of solution set(s)\operatorname{set}(\mathrm{s}) of a+b+c+d=50a+b+c+d=50.

Studdy Solution

STEP 1

1. (a,b,c,d) (a, b, c, d) is a set of integers.
2. Each integer a,b,c, a, b, c, and d d is greater than 6.
3. We need to find the number of solutions for the equation a+b+c+d=50 a + b + c + d = 50 .

STEP 2

1. Transform the problem to account for the condition that each variable is greater than 6.
2. Use the stars and bars method to find the number of non-negative integer solutions.
3. Calculate the number of solutions.

STEP 3

Transform the problem to account for the condition that each variable is greater than 6.
Since each of a,b,c, a, b, c, and d d must be greater than 6, we can set:
a=a+7, a = a' + 7, b=b+7, b = b' + 7, c=c+7, c = c' + 7, d=d+7, d = d' + 7,
where a,b,c, a', b', c', and d d' are non-negative integers.

STEP 4

Substitute these expressions into the original equation:
(a+7)+(b+7)+(c+7)+(d+7)=50. (a' + 7) + (b' + 7) + (c' + 7) + (d' + 7) = 50.
Simplify the equation:
a+b+c+d+28=50. a' + b' + c' + d' + 28 = 50.
Subtract 28 from both sides to find:
a+b+c+d=22. a' + b' + c' + d' = 22.

STEP 5

Use the stars and bars method to find the number of non-negative integer solutions to the equation a+b+c+d=22 a' + b' + c' + d' = 22 .
The formula for the number of solutions in non-negative integers of the equation x1+x2++xk=n x_1 + x_2 + \cdots + x_k = n is given by:
(n+k1k1). \binom{n + k - 1}{k - 1}.
In this case, n=22 n = 22 and k=4 k = 4 .
Calculate:
(22+4141)=(253). \binom{22 + 4 - 1}{4 - 1} = \binom{25}{3}.

STEP 6

Calculate the binomial coefficient:
(253)=25×24×233×2×1=2300. \binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300.
The number of solutions is:
2300 \boxed{2300}

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