@Estallar12 Dono ke liye @YouMadFellow is highly capable. Mast drawing bana bana ke samjhayega. 😁
@Estallar12 said:Topic I - @YouMadFellow Sir.Topic 2 - I can help under the guidance of @RoadKill Sir.
If Bf represents the number of bijective functions from S1 = {a, b, c, d, e} to S2 = {p, g, r, s, t} such that
f(c) ‰ t and f(e) ‰ p, what is the value of Bf? :P
@RoadKill said:@fisherking Ye hai kya cheez?
Agar tumne MC 1310 diya hain toh usme woh Tom and Jerry waale LR set ka solution dekhna.I think yeh Games and tournaments main sikhaya tha.
@ScareCrow28 said:If Bf represents the number of bijective functions from S1 = {a, b, c, d, e} to S2 = {p, g, r, s, t} such thatf(c) ‰ t and f(e) ‰ p, what is the value of Bf?
No of bijections = n! where n is the number of elements. Here you need to subtract the mappings where f(c)=t OR f(e)=p . Then, add back the mappings where f(c)=t AND f(e)=p. Thoda set theory type hai.
Answer should be 5!-2*4!+3! = 120-48+6=78.
@fisherking
ye lo :)
http://www.pagalguy.com/forums/cat-and-related-bschools/pg-dream-team-09-t-43726/p-1642728?page=10#post1675058
@ScareCrow28 said:If Bf represents the number of bijective functions from S1 = {a, b, c, d, e} to S2 = {p, g, r, s, t} such thatf(c) ‰ t and f(e) ‰ p, what is the value of Bf?
This is just like arranging 5 persons on 5 seats when c cannot sit in chair T and e cannot sit in chair P. :D
Total Permutations - (Those having c on T + Those having e on P) + (Those having c and e on T and P resp.). :D
=> 5! - ( 4! + 4! ) + 3! = 120 - 48 + 6 = 120 - 42 = 78 ways.
@RoadKill
Hume bhi mil gaya SBT '10 main
@Estallar12
Hume bhi mil gaya SBT '10 main
http://www.pagalguy.com/forums/cat-and-related-bschools/shout-boxers-team-2010-t-56636/p-2211314/r-2255752
@Estallar12
@Estallar12 said:This is just like arranging 5 persons on 5 seats when c cannot sit in chair T and e cannot sit in chair P.Total Permutations - (Those having c on T + Those having e on P) + (Those having c and e on T and P resp.).
If the no of elements in S2 had been 3, then? or 7? Please btado sir..concept samajh ajaega fir :P
@Estallar12 said:Two kinds of wines are mixed in the ratio 2:1 and 1:2 . The two mixtures are sold at a profit 20%and 10% respectively. If the wines are mixed in the ratio 1:1 and the individual profit percent on them are increased by 5/3 and 4/3 times respectively, then the mixture will fetch the profit
Bhai.. iska kuch batao ? Maine pehle bhi tag kiya tha ? Individual profits matlab ?
options kya hai ? 31.11 - 44.44 % ?
Aur jo kambakht bolta hain ki PG main sirf Time waste hota hain.....Usse sirf mera previous post aur Madfellow ke concept notes aur drwaings dekhne padenge.Concepts jinka kahi aur naamo nishaan nahi hain woh PG pe hain.
PG ROCKS!!! 
@ScareCrow28 said:If the no of elements in S2 had been 3, then? or 7? Please btado sir..concept samajh ajaega fir
Bhai, phir there are no bijective functions possible. Zero hoga answer I think.
@ScareCrow28 said:If the no of elements in S2 had been 3, then? or 7? Please btado sir..concept samajh ajaega fir
Bijective means one to one relation ho. No 2 images, 2 no mappings from single element.
If S2 will have 3, elements of S1 will go either without any mapping or single image from 2 values. This is not possible. Same with 7.
So, Zero hona chaiye phir.
@YouMadFellow said:Bhai, phir there are no bijective functions possible. Zero hoga answer I think.
So Sir...there has to be equal no of elements in both the sets?? Bijective means one to one mapping
@YouMadFellow said:Bhai.. iska kuch batao ? Maine pehle bhi tag kiya tha ? Individual profits matlab ?options kya hai ? 31.11 - 44.44 % ?
Ruko. Batata huun. Prep ka question hai yeh. :D
@Estallar12 Mujhe mere hi posts dhundke dega, jo tune share kiye the ? End moment me revision me kaam aayega 
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@YouMadFellow said:@Estallar12 Mujhe mere hi posts dhundke dega, jo tune share kiye the ? End moment me revision me kaam aayega !
Woh toh mere PINGs main mil jayenge & waise maine Bookmark bhi karey hain. Kisi ek din saare ek saath ek hi post main daal dunga. :D
@ScareCrow28 said:So Sir...there has to be equal no of elements in both the sets?? Bijective means one to one mapping
Arey main koi Sir nahi hu.. yeh Estallar ki baat mat suno tum. Yes, equal hone chahiye. All the elements of the domain must be mapped, for it to be a valid function, and agar one-one hai.. to har ek mapped to one each.. So, equal number of elements in both domain and co-domain or range.
@Estallar12 said:Bijective means one to one relation ho. No 2 images, 2 no mappings from single element.If S2 will have 3, elements of S1 will go either without any mapping or single image from 2 values. This is not possible. Same with 7.So, Zero hona chaiye phir.
@YouMadFellow said:Bhai, phir there are no bijective functions possible. Zero hoga answer I think.
The concept seems similar to the inverse function concept, i.e., If the inverse of a function exists that means it is a bijective function right>?
@fisherking Oh tu isey controlled sum keh raha tha kya.. Ispe article hai pg pe, padha tha pehle. But eventually question mujhse kabhi hote nahi hai exam environment mein. :(
http://www.pagalguy.com/news/cubes-matchsticks-logical-reasoning-tricks-cat-2011-a-18681
Author of this article is a KGP CS grad.. IMS mein kya kar raha hai? :O
@YouMadFellow said:Bhai.. iska kuch batao ? Maine pehle bhi tag kiya tha ? Individual profits matlab ?options kya hai ? 31.11 - 44.44 % ?
Mera 22% aa raha tha. Recheck karta huun. Shayad phir galat hoga :banghead: