Asfakul
11C8 * 8!
Had there been similar rings - - - - - - - -
= 11c8
diffrnt rings = 11c8 * 8!
@jain4444 said:How many words can be made with word "SOFTWARE" in which vowels won't be together ?
8! - 6!*3! ...
no of ways all the letters can be arranged = 8!
No. of ways when all the vowels are together = 6!*3!
no of ways when vowels are not together = 8! - 6!*3!
no of ways all the letters can be arranged = 8!
No. of ways when all the vowels are together = 6!*3!
no of ways when vowels are not together = 8! - 6!*3!
How many pair of numbers are there whose lowest common multiple is 2013?
Asfakul
4 pairs
@catahead
11! = 11*10*9*8*7*6*5*4*3*2*1 = 11*9*8*7*6*4*3*10*10 = 39916800
=> numbers ending in 8 follows a pattern of 8,4,2,6,8,4,2,6..
Since the number to the power is a multiple of 4 => 39916800^(4k) would have 6 as the last non zero digit 😃
@Asfakul
2013 = 11*183 = 11*61*3 now any two combinations of these factors would have 2013 as their lowest common multiple
1*2013, 3*671,11*183,33*61 => 4 pairs
all others would be similar to the above
4 distinct pairs...
@Asfakul
@Asfakul said:In how many ways can Mohan wear 8 distinct rings in his 4 fingers? ?
let the number of rings on each of the fingers be a,b,c,d
=> a+b+c+d=8
number of solutions to this equation = 8+4-1C4-1 = 11C3
and then the 8 rings can be arranged among themselves as well = 8! ways
=> ways = 11C3*8!
@arsharora said:@Asfakul 4 pairs!!! break 2013 into 3*11*61, then proceed!!!
Yeah I had got the same . Basically we need to find out coprime factors. Everyone else was getting some different answers so I posted it here . 
A simple albeit interesting problem.
A man earns Rs.7500. He spends 60% of his income. If his income increases by 20% and his expenditure increases by 10%, find the increase in his savings.
@VJ12 said:A simple albeit interesting problem.A man earns Rs.7500. He spends 60% of his income. If his income increases by 20% and his expenditure increases by 10%, find the increase in his savings.
35% ?
income = 7500
exp = 60*7500/100 = 4500
saving = 3000
inc. income = 120/100*7500 = 9000
inc. exp = 110/100*4500 = 4950
inc. saving = 4050
inc. in saving - 4050 - 3000/3000*100 = 1050/3000*100 = 35%
@Crysis said:@Asfakul let the number of rings on each of the fingers be a,b,c,d => a+b+c+d=8 number of solutions to this equation = 8+4-1C4-1 = 11C3and then the 8 rings can be arranged among themselves as well = 8! ways=> ways = 11C3*8!
Can we apply a+b+c+d=8 when rings are distinct ? Please pardon my ignorance . and how you are multiplying with 8! because 4 fingers can have 2,3,1,2 number of rings .
@VJ12 said:A simple albeit interesting problem.A man earns Rs.7500. He spends 60% of his income. If his income increases by 20% and his expenditure increases by 10%, find the increase in his savings.
If both income and expenditure were increased by 10%, we would have had 10% increase in savings.
What extra we have here is another 10% increase in Income. Initial Savings were 40% of Income. ie, Income was 2.5 times savings. So 10% increase in income translates to .25 savings that is 25% of initial savings.
So total increase in savings = 10% + 25% = 35%
@Asfakul said:Can we apply a+b+c+d=8 when rings are distinct ? Please pardon my ignorance . and how you are multiplying with 8! because 4 fingers can have 2,3,1,2 number of rings .
What a + b + c + d = 8 gives you is total no. of distribution of rings on the 4 fingers. At this point, we are not taking into consideration that the rings are distinct. So if the rings were similar, our answer would have been total non-negative solns to the above equation.
Once we have that number, which is C(11,3), we want to take into consideration the fact that rings are distinct. So for each of the distributions we got from the equation a+b+c+d =8, we have 8! arrangements of the rings amongs themselves.
Alternate solution can be = 4*5*6*7*8*9*10*11
Here first ring has 4 choices of fingers. 2nd ring has 5 choices in that it has 2 possibilities for the finger in which the previous ring went (below the 1st ring or above the 1st ring). so on.
If n is the largest possible value of points in a plane that any three of them can form a right triangle then what is the value of n?
plz help me out with the above...
@Angadbir said:What a + b + c + d = 8 gives you is total no. of distribution of rings on the 4 fingers. At this point, we are not taking into consideration that the rings are distinct. So if the rings were similar, our answer would have been total non-negative solns to the above equation.Once we have that number, which is C(11,3), we want to take into consideration the fact that rings are distinct. So for each of the distributions we got from the equation a+b+c+d =8, we have 8! arrangements of the rings amongs themselves.Alternate solution can be = 4*5*6*7*8*9*10*11Here first ring has 4 choices of fingers. 2nd ring has 5 choices in that it has 2 possibilities for the finger in which the previous ring went (below the 1st ring or above the 1st ring). so on.
i think there is some flaw in it
see the explanation given for similar kind of problem on this page
http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-88456/p-3611604/r-5000569?page=1465
see the explanation given for similar kind of problem on this page
http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-88456/p-3611604/r-5000569?page=1465
@hedonistajay said:i think there is some flaw in it see the explanation given for similar kind of problem on this page http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-88456/p-3611604/r-5000569?page=1465
You are confusing the problems. They are not similar.
When balls go to a person, it does not matter what order the person has the balls in.
When rings go into a finger, the order matters.