If two integers x and y such xy are selected at random between -8 and

Question

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

quantumliner · Answer

If X is -8, then the possible values for y are (-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11)

So there are 19 combinations possible, when x is -8

Similarly, if x is -7, then y can take 18 values.

So the total number of combinations = 19+18+17+16+15+..+3+2+1 = 19/2 * (19+1) = 19 * 10 = 190

Answer is c

MathRevolution · Answer

==> Since two integers from -8 to 11 are being randomly selected and x>y, you use combination. Thus, the number of integers from -8 to 11 becomes 11-(-8)+1=20, so 20, then 20C2=(20)(19)/2!=190.

The answer is C. 
Answer: C

nks2611 · Answer

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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gmatsnow · Answer

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?

SaquibHGMATWhiz · Answer

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	imes 20}{2}=190

Hence the right answer is    Option C

luisdicampo · Answer

Deconstructing the Question  
Range: Integers between -8 and 11, inclusive.
Condition: Two integers  x  and  y  are selected such that  x > y .
Goal: Find the number of possible cases.

Step 1: Determine the Size of the Set  
The set of integers is  \{-8, -7, ..., 11\} .
Total count ( n ) =  	ext{High} - 	ext{Low} + 1 
 n = 11 - (-8) + 1 = 20  integers.

Step 2: Logic of Selection  
The condition  x > y  implies that  x  and  y  must be distinct.
When we choose any 2 distinct numbers from the set, one will automatically be larger than the other. There is only  one way  to assign them to  x  and  y  to satisfy  x > y  (the larger becomes  x , the smaller becomes  y ).

Thus, the problem is equivalent to asking: "How many ways can we choose 2 distinct numbers from a set of 20?"

Step 3: Calculate Combinations  
We use the combination formula  nC2 :
 \binom{20}{2} = \frac{20 	imes 19}{2 	imes 1} 
 = 10 	imes 19 
 = 190 .

Answer:  C

AndyRugger · Answer

This is a sneaky combination vs permutations question.

Think of it like this: With 19 choices and 2 spaces, there are 380 ways to arrange all the numbers. Of that set, half will be eliminated because x will be less than y. So you divide by 2, and you get 190.

If two integers x and y such x>y are selected at random between -8 and

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