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The way I did it is
if X1=1 then X2 can take 10(2-11) values and X3 (3-12)can take 10 values
if X2=2 then X2 can take 9(3-11) values and X3 (4-12)can take 9 values and so on
1. time yourself 2. solve the problem on a sheet of paper 3. write down your solution here, and your time, if possible
solving under 2 mins would be great! go!
In how many ways can you select three integers x1, x2 & x3 from 1 to 12 such that x1<x2<x3? A. 220 B. 276 C. 318 D. 341 E. 385
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Well. Guys, i have given lot of time to everyone..only 2 guys replied. Not good if you other guys are looking for a 50 or 51Q
The answer indeed is A [220].
Geethu, yours was more of a guess , i think. but anyways.
Let me try using a different approach. it might just help guys to understand the problem better.
The first number can be picked in 12 ways.
The second number can be picked in 11 ways.
The third number can be picked in 10 ways.
total # of ways = 12* 11*10
BUT, there is a condition here.... x1<x2<x3 . The total # of ways contains some events which do not follow this pattern.
Lets say we pick 12 , 14, 5 , there are 3! ways to arrange these numbers.
Since ONLY ONE of them [in this case, 5,12,14] is required, we have to divide the total # of ways by 3!
answer : 12*11*10 /3! = 220 ways
questions?
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