How many different arrangements are possible to place seven different

Question

How many different arrangements are possible to place seven different books on a shelf if all three math books must be placed next to each other?

A. 120
B. 148
C. 360
D. 540
E. 720

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Avakash · Answer

Consider the math book as one group. Hence the total nos are remaing 4 books plus one group of math books

And is 5!

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ShankSouljaBoi · Answer

Maths books are not identical, hence must by multiples by 3! 
And E

Archit3110 · Answer

all three books together =  xxx *a*b*c*d ; 5! ways and xxx can be arranged in 3! ways 
total ways ; 5!*3! = 720
IMO E

ScottTargetTestPrep · Answer

We must treat the 3 math books as a single entity, so one arrangement of the books is as shown here:

-B4-B5-B6-B7

Treating the 3 math books as a single entity, we see that we have 5! ways to arrange the books. However, the 3 math books can be arranged in 3! ways. Thus, the total number of ways to arrange the books is 5! x 3! = 120 x 6 = 720.

Answer: E

altairahmad · Answer

generis   Bunuel

If the 3 math books were same, would we divide the answer by 3! ? Or the grouping i.e 4 + 1 would take care of this ? When does grouping take care and when do we divide by the number of similar items ?

Will appreciate your input.

ShankSouljaBoi · Answer

IF the three maths books were same 5! will be enough, one dos not need to multiply by 3!

Kudos, if this helped.

PolarisTestPrep · Answer

Hey guys

When you have to arrange a number of items and their order is important, these arrangements are called permutations

The way to quickly calculate permutations is to use factorials

I know I just said a lot of words that dont make sense, but lets go one at a time

Factorials are a math principle where if you have an integer n, a factorial is written out as n!

n! means that you multiply all of the integers from 1 to n

If n = 5, n! = 5 * 4 * 3 * 2 * 1

The reason this is used in permutations, where you arrange items in a certain order, can be understood with simple logic

Let's say you have just three books. How many different ways can you arrange them?

In the first position, there are three different books you can place there

After you've placed the first book, you only have two books left so there are only two possible books you can place in the second position

For the third position you have no choice, you only have one book remaining to put there

How does this work out to a factorial?

Well if there are three possibilities for the first position and two for the second and one for the last, that comes to a total of 3 * 2 * 1 = 6 possible arrangements

Look:

ABC
ACB
BAC
BCA
CAB
CBA

And this principle extends for any number of items

If you had to figure out how many ways there were to arrange 700 books in order like this, you would calculate 700! Good luck!!

And the reason I specify permutations is that this is one of two ways to arrange things that are commonly tested on the GMAT, the other being what is called combinations

The first step is to understand whether you are doing a permutations problem or a combinations problem

Permutation problems are the ones where the order in which items are placed matters, like in this one, and they are way easier because you just take the factorial of the total number of items to be arranged.

In combination problems, the order the things are placed in doesn't matter and the formula is more complex

If this question was just asking how many ways there were to arrange seven books, the answer would simply be 7!

But the twist is that three of the books must stay together. That means that this three book cluster can be treated as one unit. It's the three book unit and four other books that have to be arranged. The three-book unit could go first, last or somewhere in the middle between the other four books

That means there are essentially 5 things that can be ordered -- 4 individual books, and one 3 book unit

Using our knowledge of permutations that means there are 5! ways to order these things, right?

Yes, but keep in mind that inside that 3 math book unit, there are different ways to arrange those books also

How many different ways?

3!

3 different options for position one, 2 different options for position two, and 1 option for position three

3 * 2 * 1 = 6

So there are 5! ways to arrange the 5 things and one of those five things is the three book unit. There are 3! ways to arrange the three book unit

That means there is a total of 5!3! ways to arrange all the books in this way

5! = 5 * 4 * 3 * 2 * 1 = 120

120(3!) = 120(6) = 720

The answer is (E)

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How many different arrangements are possible to place seven different

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