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MathRevolution
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the answer is C

see we need to select two integers between -8 to 11( inclusive) that are total 20 integers there where x>y , so we can choose like this 20ways(y) * 19 ways(X) =380ways but the order does not matter here means whether we choose x first or y first the order is same so divide with 2 such as 380/2= 190 ways to select .

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MathRevolution, I'm not sure I understand this question. If X is > Y, couldn't X never be -8 and Y could never be 11 in the -8 to 11 range?

Would this not make the answer less than 190?
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MathRevolution
If two integers x and y such x>y are selected at random between -8 and 11, inclusive, how many cases are possible?

A. 150
B. 180
C. 190
D. 210
E. 240
Solution:

  • Since x>y, therefore y cannot be 11 and x cannot be -8

  • If x = 11, the possible values of y are 10, 9, 8, ...1, 0, -1, -2...-7, -8 i.e., 19 values
  • If x = 10, the possible values of y are 9, 8, ...1, 0, -1, -2...-7, -8 i.e., 18 values
  • ..
  • ..
  • If x = -7, the possible values of y is -8 i.e., 1 value

  • Total values \(=19+18+17+...+2+1=\frac{19\times 20}{2}=190\)

Hence the right answer is Option C
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