Permutations and Combinations

[pranjal215]

ya actually i calculated the answer wrong in the previous reply but i corrected the answer to 84 ...
actually both ways to appraoch the problem r correct but (n+r-1)C(r-1) approach is cerainly better as it is short but other one is a more visulaized answer .. apologies for the pevious incorrect post ..

correct answer is 84 ....
the disks may be chosen either all from same manufrer or all diff manfs .. and so on so u need to diffenciate in these cases and calcuklatye cases separately .

the 4^ 6 way wount work coz u r considering xtra cases in that approach ....
1) all diff mans 4 ways (4C1)
2)from 2 mans .. 2 mans in 4C2 ways X5 to arrange em (suppose a n b)
(AB two) then 4 remain (AAAA,BBBB,AABB,AAAB,BBBA)
3)4C3 X10 ways (10 similarly as above...)
4) (4C2+4)
total 4 +30+40+10
84 ways

[eric.segal1]

Guys i gotta doubt....

There are 6 sticks of different heights which are to be arranged in two rows of 3 each such that the sticks are in ascending order from left to right in each row, and the stick in front is always smaller than the one at the back. How many such arrangements are possible?

guys the answr is... 5 but i dont know the method to do it except making all such arangements... is there any short way?

ps: I've also posted the same post in the other P&C forum coz i dint know which one is active

[varun nakra1]

Originally Posted by eric.segal1

Guys i gotta doubt....

There are 6 sticks of different heights which are to be arranged in two rows of 3 each such that the sticks are in ascending order from left to right in each row, and the stick in front is always smaller than the one at the back. How many such arrangements are possible?

guys the answr is... 5 but i dont know the method to do it except making all such arangements... is there any short way?

ps: I've also posted the same post in the other P&C forum coz i dint know which one is active

a<b<c<d<e<f

1 2 3

4 5 6

#6 is the largest in the arrangement and #1 is the smallest..so 'a' and 'f' get fixed...Now the second largest is smaller than exactly the largest, so 'e' occupies either #3 or #5...and the second smallest is larger than exactly the smallest, so it occupies either #4 or #2..So 2*2 = 4 such choices and in each of the 4 arrangements, one can fix 'c' and 'd' in xactly one way..The answer shud be 4 only....

edited : it would be 5 coz in one of the cases, we can arrange 'c' and 'd' in two ways...itzz 5...

I dnt know a shoter method than this...itz more of an LR quesn than a maths one..!

[saravanan.mani]

Hi,
I have a question on permutation and combination.

Consider 6 red, 5 blue and 3 yellow pair of socks in a drawer. Electricity is not available in that room. How many time you need to take to get a pair of socks from each color?

Kindly help me to solve the above problem

Cheers,
Saran

[dushyant bhatia]

Originally Posted by saravanan.mani

Hi,
I have a question on permutation and combination.

Consider 6 red, 5 blue and 3 yellow pair of socks in a drawer. Electricity is not available in that room. How many time you need to take to get a pair of socks from each color?

Kindly help me to solve the above problem

Cheers,
Saran

Hi.. is the answer 26.. can't be that str8forward.. or am I missing sumthin.
Lemme know if I'm right
Cheerz