okay bro, first things first.there are a total of 2^15-2 ways of distributing those books on two shelves not 2^14.There are two ways in which you can count the distributions(combinatons not arrangements).
1:you have 15 books say book1,book2,...book15. Now book1 has 2 choices (to be placed on shelf 1 or shelf 2),now for every choice of book1,book2 also has 2 choices(i.e. if book1 is on shelf 1 book 2 can be on shelf1 or shelf2 or if book1 is on 2nd shelf then book2 can be on shelf1 or shelf2) similarly book3 has 2 choices for every choice of book1 and book2 continuing in this way the total number of distrbutions of the books is 2^15(Think on why we multiplied the choices of the books to get the answer).Now we have counted two cases that are not possible i:all fifteen books on shelf1 , ii:all fifteen books on 2nd shelf. So we remove those counts and get 2^15-2..
2.another way of counting the distributions is that we look at only one shelf say shelf1.
now we have a total of fifteen books so if shelf1 has 1 book on it,it can be any one of the fifteen books(the reason we are considering only one shelf is that we assume that the rest of the books will be on the other shelf,i.e. if shelf1 has 1 book on it then the rest 14 are on shelf2) so,we can select 1 book from total of 15 books in C(15,1) ways and place it on shelf1(the rest 14 will be on shelf2,so C(15,1) is actually number 1,14 combination),similarly shelf1 can have any two books on it out of 15,number of ways of selecting 2 books from 15 is C(15,2) continuing this way shelf1 can have 14 books on it but they can be any 14 out of 15 so we select 14 books out of 15 in C(15,14) ways and place it on shelf1. Finally total number of combinations is C(15,1)+C(15,2)+C(15,3)+..C(15,14),now you cant add these terms yet because there is a formula (binomial theorem) C(n,0)+C(n,1)+...C(n,n)=2^n you can see the sum that we get does not have the terms C(15,0) and C(15,15) so we add it and subtract it from the sum. So, total combinations = C(15,0)+C(15,1)+C(15,2)+...C(15,15)-C(15,0)-C(15,1)=2^15-2.
Okay now you have to understand the difference between combinations and arrangements to solve this question.
when we count the ways in which we can split the books on the two shelves without caring what their position is on their respective shelves then we are counting only the distributions(combinations) not the arrangements.So now lets see the ways in which we can split these books between the shelves. You can keep 1 book on shlef1 and rest 14 on shelf2(1,14 combination),then you can keep two books on shelf1 and rest 13 on shelf2(2,13 combination) continuing this way you get the 14,1 combination(14 books on shelf1,1 book on shelf2).We can only see 14 ways of splitting these books((1,14),(2,13)...(14,1)),but we know total ways in which we can split the books is 2^15-2,HOW??. Actually there are 14 unique ways of splitting these books and each way in itself is of different types.If you keep book1 on shelf1 and rest on shelf2 this is one type of 1,14 combination,if you keep book2 on shelf1 and rest 14 on shelf2 this is another type of 1,14 combination so adding all the types of unique combinations we get the total number of combinations.
Now, imagine any combination of the books lets say 2 books on shelf1(say book1 and book2) and the rest 13 on shelf2, now if i keep the two books on shelf1 as book1,book2 and rearrange the books on shelf2 i get different arrangements and if i keep the two books on shelf1 as book2,book1 and rearrange the books on shelf2 i get different arrangements,So imagine it this way you keep shelf1 books as it is and rearrange the books on shelf2 then rearrange the books on shelf1 once and rearrange the books on shelf2 in all possible ways you can, you will get differnet arrangements.In simple words for every arrangement of books on shelf1 you can arrange the books on shelf2 in all the way you can and you will get different arrangements.
NOW lets count..
so. if i keep 4 books on shelf1 and 11 on shelf2 i can arrange books on shelf1 in 4! ways and for every arrangement of the books on shlef1 you can arrange the books on shelf2 in 11! ways so total arrangements is 4!.11! but the combination 4,11 is of C(15,4) or C(15,11) types so total arrangements is = 4!.11!.C(15,11).In general if there are r books on shelf1 and 15-r books on shelf2 for every arrangement of the books on shelf1 books on shelf2 can be arranged in (15-r)! ways and the combination r,(15-r) is of C(15,r) or C(15,15-r) ways so total arrangements = sum(r!.(15-r)!.C(15,r)).....where r varies from 1 to 14.Now you know C(15,r)=15!/(15-r)!.r! hence every term in the sum becomes 15! and we have 14 terms, so answer=14.15!.
First of all why are you doing 15-1 to get 14.You mean to say "total no.of books - least no.of books that will be on any shelf",this does not mean anything.First decide what do you want to count is it the combination or the arrangement,Then if you want to count combinations start by knowing what combination means then come up with a systematic way to do that, do the same for arrangements.
1.Rule of sum or product? or is it C(n,r) or P(n,r)??(we always add):
First,i suggest you to find the meaning of every formula and rule,like P(n,r) means "adding all the arrangements of r objects selected from n distinct objects".So understand the result of each and every formula and rule.
Now if you are confused when to use rule of sum or rule of product or when to use C(n,r) and when do i use P(n,r), then know that actually we only add,we use these rules to fasten our calculations (same is for all the formulas i.e sum(C(n,r)) r varies from 0 to n = 2^n,i.e don't directly jump onto the formulas use them to fasten your calculations).Now
its all about that case..Now, if you have 10 books and you are asked if you can select one book out of it ,in how many ways can you do that.The answer is 10,because case 1 you pick book1,case 2 you pick book2 so if you add up all the cases you get the answer.Now imagine if you are asked to count the total fingers you have,you can count your fingers one by one OR you say i have same fingers on both hands so i count the fingers on one hand and multiply it by 2(The rule of product was used to fasten our calculation.Also whenever you see multiplication always see it as "a certain number of things for every thing").Now consider this question,the first case is that you split the books in 1,14 combination but this case itself is of 15 types,now to count case 1: first we have to count the arrangements,how do we do that ?, we come up with a way of imagining those arrangements(i.e keeping books on shelf1 as it is then arranging the books on shelf2 in every way we can,then arrange the books on shelf1 once and arrange the books on shelf2 in every way we can.),so when we imagine all the arrangements we realize that we need product rule and the P(n,r) to fasten our calculations,then every type of 1,14 combination has the same no.of arrangements so instead of adding we will just use rule of product,Similarly we count all the different cases and add them up.
Whenever you approach a counting problem,divide it into cases count each case in a systematic way by imagining what it will look like,then add up all the cases.Remember to always add and use the formulas to only fasten your calculations,don't directly jump on them.
Finally there may be times when you come up with a wrong way of counting,then don't simply discard that method rather justify yourself why it is wrong.