Distribution of Alike Objects

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10 Comments

Aditya on October 26, 2013 at 6:27 pm
Thanks a lot 🙂
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Neha on June 19, 2013 at 8:55 pm
Dude you guys are just awesome..!!! cant expect anything better..
a great site with great content..
don’t know why is it not that famous…
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SA on September 5, 2012 at 7:36 pm
Very Good explanations … Thank you so much..
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hiccup on May 20, 2012 at 8:46 am
@prashant,
nice article.
how to go about finding number of distinct terms in the expansion (1+x^5+x^10+x^15)^35
is it also somewhat related to this concept??
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- Neha on June 19, 2013 at 8:59 pm
  hey hiccup..
  this is same as above concept…
  you have to take the number of terms inside brackets.. no matter what the term is just take the number…
  weather it is a,b,c,d or (1+x^5+x^10+x^15).. doesn’t matter..
  so here the answer will be —> ” 37! / (3! * 34!) “
Iconoclast on April 26, 2012 at 10:30 am
Again, thanks for such a great article…
The best part of the article is the way it moves from simplest to the toughest level.
I took the printout of your first article yesterday and when I finished it, I felt very confident because of the depth of the topic covered in that.
I find same thing in this article too. I hope new articles are posted more frequently.
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Amit Chhabra on April 26, 2012 at 7:22 am
Good article ..
Kudos to Puneet!
But some typo errors were there in
Ques 6 : End part (solution contains figures of ques 5)
Ques 8 : equation above equation 1 should be 2k+ y+ z =18
Ques 11 : Solution should be 22!/ (19! * 3!)
Waiting eagerly for another article(on P&C and probability)
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- prashant on April 26, 2012 at 9:16 am
  ahhh
  thanks alot amit…the errors have been corrected….
  and yeah more posts coming on the topic soon..:)
Niteesh on April 25, 2012 at 2:24 pm
Very good one..!!!!
Reply
- prashant on April 25, 2012 at 2:34 pm
  thanks alot niteesh…:)

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