There are 100 different books in a shelf. Number of ways in which 3 bo

Question

There are 100 different books in a shelf. Number of ways in which 3 books can be selected so that no two of which are adjacent is

(A) 100C3 – 98 
(B) 97C3
(C) 98C3 
(D) 96C3
(D) 95C3

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ScottTargetTestPrep · Accepted Answer

Solution:

We can solve this problem by using a much smaller number for the total number of books on the shelf. For example, if there are only a total of 5 books on the shelf, say, A, B, C, D, and E (and assuming the books are on the shelf in that order), then there is only 1 way to choose the 3 books such that no two of them are (originally) adjacent (to each other): A-C-E.

If there are only a total of 6 books on the shelf, say, A, B, C, D, E, and F (and assuming the books are on the shelf in that order), then there are 4 ways to choose the 3 books such that no two of them are (originally) adjacent (to each other): A-C-E, A-C-F, A-D-F, and B-D-F.

So let’s say n is the total number of books on the shelf. Using these two rather easy examples, we see that when n = 5, the total number of ways to choose three books without any two of them being adjacent is 1, which is 3C3, or (5 - 2)C3, and when n = 6, that total is 4, which is 4C3, or (6 - 2)C3. Therefore, when n = 100, the total will be (100 - 2)C3 = 98C3.

Answer: C

nick1816 · Answer

a B1 b B2 c B3 d

a+b+c+d=97

where a and d are non-negative integers and b and c are positive integers

a+1+B+1+C+d=97, where B and C are non-negative integers

a+B+C+d=95

a|B|C|d

We have to arrange 95 similar objects and 3 partitions.

Total possible solutions = (95+3)!/95!3! = 98C3

Fdambro294 · Answer

Let’s see if I can do this explanation any justice....

Imagine the 100 books in a row.

For the first selected book, there does not have to be any books before this selection. We’ll call the “GAP” before selection 1 A.

There can be 0 books in Gap A (meaning we select the 1st book on the shelf)

There can be 1 book in Gap A (meaning we select the 2nd book on the shelf)

There can be 2 books in Gap A (meaning we select the 3rd book on the shelf)

No before we make the 2nd selection, we need to have another Gap of at least 1 book in between Selection 1 and selection 2. This is because no 2 books selected can be adjacent. We can call this Gap B.

Gap B must have > or = 1 book. In other words, there must be at least 1 book separating selection 1 and selection 2.

Similarly, there needs to be another Gap (call it Gap C) between selection 2 and selection 3 that contains > or = 1 book. In other words, there must be at least 1 book separating selections 2 and 3.

Finally, after selection 3 there can be another Gap (call it Gap D). Similar to Gap A, there can be 0 books in Gap D following selection 3.

There can be 0 books after selection 3 (meaning selection 3 is the 100th book)

There can be 1 book after selection 3 (meaning selection 3 is the 99th book).

At this point it becomes a standard “Stars and Bars” type problem:

A (selection 1) B (selection 2) C (selection 3) D ——— (100 - 3 books selected = 97 books to distribute among the “gaps”)

Assuming the 97 books are like identical chocolates, we need to find all the arrangements of 97 identical chocolates and 3 partitions to determine how many different arrangements with books in and around the selections.

As a linear equation:

A + B + C + D = 97

In which B and C can be any integer that is 1 or greater

and A and D can equal any non negative integer (0 or greater)

This is similar to having to distribute 97 identical chocolates to 4 children (here the children are the “gaps”) in which:

A can receive 0 or more

B must receive at least 1

C must receive at least 1

D can receive 0 or more

After giving the 1 chocolate (the 1 book separating the selections) to B and C, we have:

A + b + c + D = 95

Solving using the partition method or stars and bars method or whichever name you want to give it, we end up with:

(95 + 3)! / 3! * 95!

98! / 3! * 95! = which is the same as “98 choose 3”

Answer (C)

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Fdambro294 · Answer

I wanted to leave this post separate.

I believe this is a concept that shows up on the Indian CAT (in a much more complex manner).

Considering we only get a couple (maybe) P and C questions, I would stay away from this question (but I’m not an expert lol).

If this were to pop on the exam,  ScottTargetTestPrep  solution would be the way to go.....thinking on your feet.

Posted from my mobile device

Regor60 · Answer

Consider a vertical bar as one selected book.

Consider a space between vertical bars as containing at least one unselected book.

Consider a space outside the vertical bars as containing 0 or more books.

Represent this as

_|_|_|_

Since the 3 vertical bars represent selected books, there are now 97 books left to fill the spaces.

If you were to then allocate the remaining books without constraint among the spaces, the two spaces between the 3 selected books could have 0, which would violate the noncontiguous condition of the question.

So, first allocate 1 book to each of those two spaces to take care of this.

This leaves 95 books to allocate at will.

Using the separator method, there are 3 separators plus 95 identical objects for 98 objects to be arranged.

98! arranges these objects.

Then dividing by 3! and 95! adjusts for the lack of distinction between the bars and books, respectively, or

98!/3!95!

Posted from my mobile device

bumpbot · Answer

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