The set S is defined as {1, -2, -3, 8, 21, 30, -22, -5, 6, -11}. The chi of set is defined as the product of any 3 elements of the given set S. How many values of chi are possible such that chi is at least 7?
A. 50
B. 59
C. 60
D. 61
E. 120
OA B
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The set S is defined as {1, -2, -3, 8, 21, 30, -22, -5, 6, -
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The wording doesn't make sense, because they're saying chi is "defined" in a certain way, and chi is not well-defined. What is the chi of this set? Who knows, there are dozens of possible values.
Anyway, if we need a value greater than 7, we certainly need the value to be positive. The set contains 5 positive and 5 negative numbers. If we'll pick three of them, and the product will be positive, we must either pick 3 positive numbers, or 1 positive and 2 negative numbers. We can pick 3 positive numbers in 5C3 = (5)(4)/2! = 10 ways. We can pick one positive and two negatives in 5C1*5C2 ways, or (5)*(5)(4)(3)/3! = 50 ways. So in total, there are 10+50 = 60 ways to get a positive product. We then need to rule out any positive products that will be less than 7, but we only get a positive product less than 7 if we choose specifically the values 1, -2 and -3, so in only one way, leaving 59 positive products greater than 7. There is one last thing to check - that you can't get equal values of the product in two different ways - but glancing at the set, that doesn't appear to be possible.
Anyway, if we need a value greater than 7, we certainly need the value to be positive. The set contains 5 positive and 5 negative numbers. If we'll pick three of them, and the product will be positive, we must either pick 3 positive numbers, or 1 positive and 2 negative numbers. We can pick 3 positive numbers in 5C3 = (5)(4)/2! = 10 ways. We can pick one positive and two negatives in 5C1*5C2 ways, or (5)*(5)(4)(3)/3! = 50 ways. So in total, there are 10+50 = 60 ways to get a positive product. We then need to rule out any positive products that will be less than 7, but we only get a positive product less than 7 if we choose specifically the values 1, -2 and -3, so in only one way, leaving 59 positive products greater than 7. There is one last thing to check - that you can't get equal values of the product in two different ways - but glancing at the set, that doesn't appear to be possible.
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We have that
\( \begin{cases}
+\text{ve integers} = 5 \\
-\text{ve integers} = 5
\end{cases}
\Rightarrow 10 \text{ integers}\)
Tota chi possible number i.e product of 3; \(10c3 = 120\)
Now, cases where the number is not \(\leq 7\)
\( \begin{cases}
\text{all -ve integers} = 5c3 = 10 \\
2\text{ +ve and } 1 \text{ -ve}; 5c2\cdot 5c1; 10\cdot 5 = 50\\
1\text{ +ve and } 2 \text{ -ve}; (1, -2, -3); \text{only } 1 \text{ case} = 1
\end{cases}\)
Total number of integers \(\leq 7 = 10 + 50 + 1 = 61\)
So, chi is not \(\geq 7 \,\Rightarrow\,120-61=59\,\Rightarrow\) __B__.
\( \begin{cases}
+\text{ve integers} = 5 \\
-\text{ve integers} = 5
\end{cases}
\Rightarrow 10 \text{ integers}\)
Tota chi possible number i.e product of 3; \(10c3 = 120\)
Now, cases where the number is not \(\leq 7\)
\( \begin{cases}
\text{all -ve integers} = 5c3 = 10 \\
2\text{ +ve and } 1 \text{ -ve}; 5c2\cdot 5c1; 10\cdot 5 = 50\\
1\text{ +ve and } 2 \text{ -ve}; (1, -2, -3); \text{only } 1 \text{ case} = 1
\end{cases}\)
Total number of integers \(\leq 7 = 10 + 50 + 1 = 61\)
So, chi is not \(\geq 7 \,\Rightarrow\,120-61=59\,\Rightarrow\) __B__.
