I dont see any difference between the two cases you have shown.
Case 2.
You need to find out the number of ways of choosing two numbers satisfying the criteria given.
Using the formulae suppose u got the answer as "P"
Case 1
Here you have to find out the ordered pairs of the same. So it will evidently be the double of Case 2 ((a,b) and (b,a) would have considered to be same in case 2 where as they have to be considered as different pairs in case 1) . and the answer will be "2P"
Am i speaking sense here? Or by any chance you were looking out for the logic of deriving the formulae?
Hi guys............hope every1's prep is going well.wasn't active on pagalguy the lat few days, bcoz i had gone on a tour with my students.Just to let all of u know,The IIMK admission criterion for the next academic year is available on the TIME website
@Kevin88 said:@seetharam7 No problem brother !!@raku1989 Here's a Q through which you can explain for the benefit of allFind the no of unordered pairs (a,b) yielding an LCM of 720. (Pls post your steps too!!)Hope everyone's prep is goin well @rohan_bhaskar Brother what ever happened to our grand plans of getting this thread proactive ?
Is the answer 120? if its mentioned in the qn that "a" and "b" are distinct then answer is 119.
720 = 2^4*3^2*5
Let a = 2^x1*3^y1*5^z1
b= 2^x2*3^y2*5^z2.
Now just consider the power of 2. in the LCM power has to be 2^4.
So in either a or b power has to be 2^4.
If you consider the power of 2 in "a" is 4, then power of 2 in b can take values from 0 to 4.
ie a total of 5 values.
Same way if you consider power of 2 in "b" is 4, then x1 can also take 5 values.
So a total of 10 values is possible for 2 .....(2*(4+1))
Do the same for the power of 3. You will get a total of 6 values. (2*(2+1))
For the power of 5 ull get 4 values. (2*(1+1))
So the total number of ordered pair is 10*6*4 = 240 pairs.
Out of these 240 pairs there will be two equal pairs we would have considered where a=b=720.
So for the un ordered pairs if a and b are distinct then ans is (240-2)/2= 119.
If a and b can be equal then ans is 240/2 = 120
Correct me if i am wrong somewhere. Hope my explanation is clear enough.
There seems to be a mistake here, its actually only 9 values as u would have considered the case of a=4 & b=4 twice.So total ways = 2(4)+1 =9.Now proceeding on a similar manner : Total no of ordered pairs = 9*5*3 = 135 pairs.This is from where i have a doubt :Is the No of unordered pairs = {(135-1)/2} = 67 ???Correct me if I'm wrong!! I don't have an answer, as i just thought about it during work today
Consider a= 2^4*3^2*5^0.
b=2^4*3*5^1.
here a=b=4....and still LCM is 720.
So its not that i considered the same entities twice.
overall there will be only one case where we consider the same case twice. when a=b. ie at a=2^4*3^2*5^1 and b also = 2^4*3^2*5^1. this particular case we will subtract.
I still feel its 2*(4+1) = 10 values. not 9. what say?
