CA Animesh

CAT Preparation Community 7y ago

Hi Team  My Doubt is conceptual in nature invoving more than two concept ,that is why i am asking here . Please help get clarity on this , i would be extremely grateful .Please be assured that i have spent entire day podering over this and when i did not get it now at the end of the day i am troubling the team. Doubt 1 :  What is difference between two formula for partition , "  r^n  " and "  (n+r-1)! / (r-1)!  " ? I believe that where arrangement is also required then later formula is used . Example : 5 Distinct Chocolate is distributed to 3 children , and it also matters order in which these chocolate is received by children then  "  (n+r-1)! / (r-1)!  " will be used. And if it  matters in how many ways children can get chocolate, no need for any order then "  r^n  " is used . Please confirm whether i am right in my logic  Doubt 2 : What is purpose of diving by  " (r-1)!  " in formula   (n+r-1)C(r-1) and formula  (n+r-1)!/(r-1)! I could not think it through , my only guess is to remove duplicacy but i dont know how that duplicacy is formed .by (n+r-1)! I know one logic explained in video that in case of repitition  (in example where 0 is chocolate and 1 is r ,i.e 00000100010000 ) , (r-1)  is represented by 1 and its repitative so to remove it ,we divide by (r-1)! as we do for repitition .  But we can see in formula theory that   "r" is "DISTINCT"  means r is slots or boxes or children which are distinct so they can not be duplicate .  Why is Formula divided by (r-1)! . Thanks a ton for your time , waiting to hear from you soon. Just last thing , my friend tells me i am putting unnecessary thoughts by diving  deep into how formula works . Please advice whether i am wasting my time as it is very time consuming to dig deep to roots and i am CA so no mathematical background after 12th so it is not possible for me to grasp working of formula of many topics as algebra and Number system  I will be very grateful for your patience and time  Animsh Kumar

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Maruti KonduriFaculty 7y ago

Suppose there are three children A, B and C and we have to give them one fruit each. There are 5 varieties of fruits -> P, Q, R, S, T and U If the number of fruits of each variety is large. The number of ways of doing this is 5^3 as each child can get a fruit of any of the 5 varieties. So, total will be 5*5*5 = 125 If there is exactly one fruit of each variety, the number of ways of distributing will be 5*4*3. The first child can get any of the five varieties, the second child will get any of the remaining 4 and the last child will get the any of the remaining three.

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