The number of ways in which a student can make selection of one or more books is ?

sorry nishant sir ,i didnt get it plz explain me

i look into the problem like this....... a student can either choose a book or he can reject it..... so there is types of 2 fates for each book...and there are are k books.....l copies of each.....so total kl books........ so total no of selection is 2^kl...but that includes choosing no book..... so the total probable ways of choosing one or more than one book is 2^kl-1..... but this doesn't match with the answer.....so where had i gone wrong??????plz help out.....

that because you are distinguishing between copies of the same book.

Supposing there are three copies of a book. In a particular selection if two copies of this book are chosen, it does not matter which two are taken.

yes nw got it..............

dats the apple mango n orange problem....which dis post reminds me of

if there are m aples..n mangoes n p oranges...
how many ways can u chose atleast 1 fruit???

yeah. that, in fact is the generating function approach.

If you look at the function

(1+x+x²+...+x^k)((1+x+x²+...+x^k))...(1+x+x²+...+x^k) (l such products)

and expand it out as a₀+a₁x+...+a_kla^kl, then the coefficient of x^r gives you the number of ways of selecting r books.

So we have to find a₁+...+a_kl (remember at least one book has to be chosen)

a₀=1.

Now put x = 1.

LHS = (k+1)^l

RHS = a₀+a₁+...+a_kl

= 1 + a₁+...+a_kl

Hence a₁+...+a_kl = (k+1)^l-1

Thank you
:-)