@Subhashdec2 said:bhai men ka arrangement 3!*2 ways mein nahi hoga?
Nai bhai..kyu?? Simple arrangements of 4 diff objects around a circle = (n-1)! he hoga na
@Buck.up said:4 couples(4 women and 4 men) are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.Find the number of arrangements.
12 ways ?
3!*2 = 12 ways
3! for 4 men...and 4 women can sit around the table in 2 ways only
@ScareCrow28 said:Nai bhai..kyu?? Simple arrangements of 4 diff objects around a circle = (n-1)! he hoga na
abcdefgh
either men can be placed on seats aceg or bdfh
3! in both cases
3!*2
@Subhashdec2 said:abcdefgheither men can be placed on seats aceg or bdfh3! in both cases3!*2
I don't think there is any distinction in the seats..they are all the same..just the position relative to each other matters.. 
Otherwise question would have stated clearly..Generally we do like this or else there will be no such thing as (n-1)!
Otherwise question would have stated clearly..Generally we do like this or else there will be no such thing as (n-1)!@Buck.up said:Yes. I think. This is right.Women arrangement should be 4 ways ? But see chill saar solution( Post no #256). Even spectramind got same 12.http://www.pagalguy.com/forums/quantitative-ability-and-di/permutations-and-combinations-t-4826/p-91145?page=13Are we doing some wrong ?
12 hi ho ga yar arrangement bna k dekho 1 arrangement cancel ho gyi 12,,,, did it last year still wrong screww me@!!!!!!!!!!!!!!!!!
@Subhashdec2 said:abcdefgheither men can be placed on seats aceg or bdfh3! in both cases3!*2
This is only if the seats are distinct to begin with. In which case anyway it will be 4! x 2 and not 3! x 2 ?
regards
scrabbler
regards
scrabbler
@abhishek.2011 said:12 hi ho ga yar arrangement bna k dekho 1 arrangement cancel ho gyi 12,,,, did it last year still wrong screww me@!!!!!!!!!!!!!!!!!
And I thought only I have that problem
@Subhashdec2 @ScareCrow28 @abhishek.2011
Yes. I got it. diagram banaya fir samjha. Women can be arranged in 2 ways only
@sukhada said:CAT 2008 Number System question if anyone is interested :Numbers 1 to 40 are written on board. keep erasing any two Numbers, say (a,b) and replace by (a+b-1). Exactly one number will remain after some time..what is that?
sequence n(n+1)/2 +1
put n =39
781 ???
When writing a math expression, any time there is an open bracket "(", it is eventually followed by a closed bracket ")". When we have a complicated expression, there may be several brackets nested amongst each other, such as in the expression (x+1)∗((x−2)+3(x−4)×(x^2+7×(3x+4))). If we removed all the symbols other than the brackers from the expression, we would be left with the arrangement ()(()()(())). For any arrangement of brackets, it could have come from a valid mathematical expression if and only if for every place in the sequence, the number of open brackets before that place is at least as large as the number of closed brackets. If 34 open brackets and 34 closed brackets are randomly arranged, the probability that the resulting arrangement could have come from a valid mathematical expression can be expressed as a/b wherea and b are coprime positive integers. What is the value of a+b?
@Tusharrr said:When writing a math expression, any time there is an open bracket "(", it is eventually followed by a closed bracket ")". When we have a complicated expression, there may be several brackets nested amongst each other, such as
@chillfactor @scrabbler this problem can be done by recursion?
@sukhada said:CAT 2008 Number System question if anyone is interested :Numbers 1 to 40 are written on board. keep erasing any two Numbers, say (a,b) and replace by (a+b-1). Exactly one number will remain after some time..what is that?
I do such questions but taking smaller values instead of 40 and the logic becomes very clear
@catahead said:I do such questions but taking smaller values instead of 40 and the logic becomes very clear
(1+2+3+.....+40)-39*1=781
@albiesriram said:@chillfactor @scrabbler this problem can be done by recursion?
Yeah but the number would be huge by the time we come to 34 😲 Have no idea how this is supposed to be solved...this is an olympiad style problem.
Was doing a little trial and error, each time number of ways comes to *3 -1 from the previous shayad.
Brkts #ways
1 1
2 2
3 5
4 14 and so on. Might have made some mistake though.
@Tusharrr 2 requests please:
(a) Let me know where you are posting these problems from. In the last 3 days you have posted several problems which are way over CAT level - even JEE does not go to this depth.
(b) Post OAs and solutions as well as just posting problems, else we have no way to know whether our answers/approaches are right...I have already requested you to do this before in a PM, which you acknowledged, but you have still not answered any of the earlier ones either.
regards
scrabbler
Was doing a little trial and error, each time number of ways comes to *3 -1 from the previous shayad.
Brkts #ways
1 1
2 2
3 5
4 14 and so on. Might have made some mistake though.
@Tusharrr 2 requests please:
(a) Let me know where you are posting these problems from. In the last 3 days you have posted several problems which are way over CAT level - even JEE does not go to this depth.
(b) Post OAs and solutions as well as just posting problems, else we have no way to know whether our answers/approaches are right...I have already requested you to do this before in a PM, which you acknowledged, but you have still not answered any of the earlier ones either.
regards
scrabbler
@scrabbler said:Yeah but the number would be huge by the time we come to 34 Have no idea how this is supposed to be solved...this is an olympiad style problem.
I was also searching for some patterns.. . .
. .Recursion mein recursion use karne ka technique hi kyaa? 
I have read one pattern to solve this bracket wala sums before cat.
it is similar to forming hills with n upstrokes and n down strokes i suppose ..
for example
/\
/ \ / \ likewise.. i forgot the pattern somehow :(
Edit:-
Grilled my grey matter to recover the pattern. working on. will try to post the OA
@Buck.up said:4 couples(4 women and 4 men) are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.Find the number of arrangements.
circle => mark 8 positions , now select 4 men arrange dem in 4 positions ( leaving one space ) ==> 3! ways ( circular arrangement ) .
Now 4 gaps are left , say first gap hav M1 _ M2 _ M3 _ M4 ==> four womens can be placed in 2 ways .
Total ways = 6*2 = 12?
Now 4 gaps are left , say first gap hav M1 _ M2 _ M3 _ M4 ==> four womens can be placed in 2 ways .
Total ways = 6*2 = 12?
@abhishek.2011 said:Find the least number of all equal weight pieces which can be cut from three cakes of weights; 484g, 308g and 253g.A. 90B. 95C. 100D. 105
HCF ( 484-308 ,308-253 ) = HCF ( 176 , 55 ) = 11 ;
total pieces = 484/11 + 308 /11 + 253/11 = 44 + 28 + 23 = 95
total pieces = 484/11 + 308 /11 + 253/11 = 44 + 28 + 23 = 95
@Tusharrr said:When writing a math expression, any time there is an open bracket "(", it is eventually followed by a closed bracket ")". When we have a complicated
Its basically an application of catalan numbers
Total arrangements 68 !/ 34!*34 !
Reqd valid arrangements = 1/ n+1 * C(2n,n)
Here,
n = 34
wolfram gives the answer as 1/35 , so a+b = 36 may be wrong :(
for more refer this.