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

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If two integers x and y such x>y are selected at random between -8 and [#permalink]

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New post 16 May 2017, 01:33
2
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A
B
C
D
E

Difficulty:

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Question Stats:

75% (01:19) correct 25% (01:38) wrong based on 59 sessions

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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
Spoiler: OA

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Re: If two integers x and y such x>y are selected at random between -8 and [#permalink]

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New post 16 May 2017, 14:48
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
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Re: If two integers x and y such x>y are selected at random between -8 and [#permalink]

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New post 18 May 2017, 01:08
==> 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
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Re: If two integers x and y such x>y are selected at random between -8 and [#permalink]

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New post 22 May 2017, 03:24
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 .

if you liked it press kudos
Re: If two integers x and y such x>y are selected at random between -8 and   [#permalink] 22 May 2017, 03:24
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