There are 10 books on a shelf, of which 4 are paperbacks and 6 are har

Question

There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A) 75
B) 120
C) 210
D) 246
E) 252

chetan2u · Accepted Answer

the best way would be..

1) choosing ways where all 5 are of one kind and subtracting from all ways...
total ways-  10C5 
all one of a kind- 6C5 
total selections= \frac{10!}{5!5!}-\frac{6!}{5!1!}=246 
D

bhaskar438 · Answer

You can calculate directly possible selections of 5 books from the shelf that contain at least one paperback and at least one hardback directly as  shyind  did or indirectly as  chetan2u  did.

I will elaborate further on chetan2u's method. To calculate the possible selections of 5 books from the shelf that contain at least one paperback and at least one hardback, subtract the complement of at least one paperback and at least one hardback from total number of 5 book combinations. To find the complement event, use DeMorgan's rule.

DeMorgan's Rule  states that ~(P and Q) = ~P or ~Q
~ symbol means  Not

So if we let  P  = at least one paperback then  ~P  = No paperbacks.
So if we let  Q  = at least one hardback then  ~Q  = No hardbacks.

No hardbacks would mean 5 paperbacks, but this is not possible since there are only 4 paperbacks. ~Q = 0 ways
No paperbacks would mean 5 hardbacks, which can be chosen in  C^6_5  in 6 ways. ~P = 6 ways
~P or ~Q = ~P + ~Q = 6 + 0 = 6 ways

Total number of 5 book combinations from 10 books is  C^{10}_5  =  \frac{(10*9*8*7*6)}{(5*4*3*2*1)}  = 252 ways

(Total number of 5 book combinations) - (No paperbacks or No hardbacks) = 252 - 6 = 246.

Correct answer choice is  D.

shyind · Answer

Paperbacks - 4, hardbacks - 6

5 books in total and at least 1 from each.

Total combinations for 5 books = (1pb, 4 hb) + (4pb, 1hb) + (3pb, 2hb) + (2pb, 3hb)

1pb, 4hb = 4c1*6c4 = 60
4pb,1hb = 4c4*6c1 = 6
3pb,2hb = 4c3*6c2 = 60 
2pb,3hb = 4c2*6c3 = 120

Total combinations of 5 books = 60+6+60+120 = >246

Ans D.

adiagr · Answer

Let us have no restrictions and select 5 books out of available 10 books.

This can be done in 10C5 ways =  \frac{10!}{5!5!}  = 252 ways.

Now we need to take out those cases when all books are hardbacks or all books are paperbacks.

Note that when we select all paperbacks, there are only 4 options. So in any case we have to choose one hardback. Condition is satisfied.

So basically we need to subtract only those cases when all hardbacks are chosen.

5 books can be selected out of 6 hardbacks 6C5 ways =  \frac{6!}{5!1!}  = 6 ways.

Required answer: 252- 6 = 246 ways

D is the answer.

ScottTargetTestPrep · Answer

First let’s determine the total number of ways to select 5 books from 10 books:

10C5 = (10 x 9 x 8 x 7 x 6)/5! = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 3 x 2 x 7 x 6 = 252

Since we are selecting 5 books and there are only 4 paperbacks in this 5-book selection, there must be at least 1 hardback. On the other hand, the only way that no paperbacks are in the 5-book selection is when all 5 books are hardback. Since we also want to have at least 1 paperback in the 5-book selection, we need to subtract the number of ways that all 5 hardback books are selected from 252, the total number of ways 5 books can be selected from 10 books.

Number of ways 5 paperback books are selected is:

6C5 = 6

Thus, the total number of ways to select the books with at least 1 paperback and 1 hardback is 252 - 6 = 246 ways.

Answer: D

Bunuel · Answer

There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A. 75
B. 120
C. 210
D. 246
E. 252

You will get an answer with duplications in it.

We have 4 paperback and 6 hardback books:
p1, p2, p3, p4, h1, h2, h3, h4, h5, and h6.

One of the cases possible with 4*6*8C3 would be p1 (from  4 *6*8C3), h1 (from 4* 6 *8C3) and p2, p3, p4 (from 4*6* 8C3 ). So, {p1, h1, p2, p3, p4}.

Another possible case would be be p2 (from  4 *6*8C3), h1 (from 4* 6 *8C3) and p1, p3, p4 (from 4*6* 8C3 ). So, {p2, h1, p1, p3, p4}.

But those two cases are the same, which means that with this approach we are getting a lot of cases double-counted.

Hope it's clear.

Divyadisha · Answer

Ways to choose 1 paperback and 1 handbook

1 P 4H
2P 3H
3P 2H
4P 1H

Basically, there just one way NOT to select at least 1P and 1H- if one selects all 5H

Total ways of selection - ways to select all 5H will give ways to select at least 1P and 1H

10C5- 6C5= 10!/5!5! - 6!/5!= 246

D is the answer

rakeshpuls123 · Answer

emi111 
you can solve the question by using your method, but it will give you multiple counts for same event. you have to take care of that counts and subtract it. Although your method will give answer, it is lengthy and confusing and strongly not recommended.

ankur2710 · Answer

Please someone elaborate on this.

Why is this logic incorrect - 4 * 6 * 8C3 ?

BrentGMATPrepNow · Answer

Key observation: It's impossible to have a selection of 5 books, in which none of the books are hardbacks. 
In other words, there will always be at least 1 hardback book in a collection of 5 books, which means we just have to deal with having  at least one paperback

Well use to formula:  # of ways to obey the restriction = (# of ways to ignore the restriction) - (# of ways to break the restriction)

# of ways to ignore the restriction  
In other words, in how many ways can we select 5 books from 10 books?
Since the order in which we select the books does not matter, we can use combinations. 
We can select 5 books from 10 books in 10C5 ways
 10C5 = \frac{(10)(9)(8)(7)(6)}{(5)(4)(3)(2)(1)} = \frac{(9)(8)(7)(6)}{(4)(3)(1)}= \frac{(9)(2)(7)(6)}{(3)(1)}= (3)(2)(7)(6) = 252

# of ways to break the restriction  
In order to break the restriction, we must have 0 paperbacks in the selection of 5 books, 
In other words, we must select 5 books from the 6 hardbacks. 
Once again we'll use combinations. 
We can select 5 hardbacks from 6 hardbacks in 6C5 ways
 6C5 = \frac{(6)(5)(4)(3)(2)}{(5)(4)(3)(2)(1)} = 6

So, # of ways to  obey  the restriction  = 252 - 6 = 246

Answer: D

PGTLrowanhand · Answer

Hi GMATters,

Here's my video solution to this problem:

https://www.youtube.com/watch?v=AJ4yOY4dtYk

Best,

Rowan

Vegita · Answer

BrentGMATPrepNow

4C1 * 6C1 * 8C3

I understand the complementary route but if we go by this approach, is there a way to remove the duplicates?

BrentGMATPrepNow · Answer

I don't recommend that approach.

Your calculations suggest you are first choosing 1 paperback (in 4C1 ways), and 1 hardback (in 6C1 ways).
This ensures that you have one of each book type (as the question demands) 
From there, you choose 3 books from the remaining 8 (thus the 8C3)

At this point, it's VERY difficult (unless I'm missing something) to determine the various ways you have over counted certain outcomes.

rahulbiitk · Answer

Bunuel   BrentGMATPrepNow   karishmaB  - I didn't quite understand the approach itself though. I've used this multiplicative approach in other problems, and it's usually worked. Why does it break in this case? I used this and ended up answering incorrectly.

Would appreciate your insights!

MBAHOUSE · Answer

There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

A) 75
B) 120
C) 210
D) 246
E) 252

MartyMurray · Answer

That approach doesn't work because the books you choose from for 4C1 and 6C1 are from the same set from which you choose for 8C3. In other words, the other 3 books you do not choose in each case for 4C1 and the other 5 books you do not choose in each case for 6C1 are in the 8C3 set.

So, you end up with multiple ways to choose the same set of five books and thus overcount the number of possible sets of five.

egmat · Answer

This is a classic combinatorics problem that tests your ability to handle "at least" constraints efficiently. Let me walk you through the key strategic insight. 
The Strategic Breakthrough:
Instead of trying to directly count all valid combinations (which would require analyzing multiple cases like "1 paperback + 4 hardbacks," "2 paperbacks + 3 hardbacks," etc.), we use complementary counting.
 Step-by-Step Approach: 
 
 Total unrestricted selections:  C(10,5) = \frac{10!}{5! 	imes 5!} = \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 252  
 Invalid selections (violating our constraint):  
 
 All paperbacks only:  C(4,5) = 0  (impossible - can't choose 5 from 4) 
 All hardbacks only:  C(6,5) = 6   
 
 Final calculation:  252 - 0 - 6 = 246   
 Final Answer:  \boxed{D) 246}

Why This Approach Works:
- The beauty of complementary counting is that it transforms a complex multi-case problem into simple subtraction. Rather than getting lost in "how many ways can I pick exactly 2 paperbacks and 3 hardbacks, then 3 paperbacks and 2 hardbacks..." we simply ask: "What are the ways that DON'T work?"
- Critical Recognition: When you see "at least one of each type" in combinatorics, immediately consider whether complementary counting might be cleaner than direct counting.

For the complete breakdown including common trap analysis and how this pattern appears across similar constraint problems:  https://neuron.e-gmat.com/quant/questions/there-are-10-books-on-a-shelf-of-which-4-are-paperbacks-and-6-are-har-1782.html

ADisHere · Answer

the best way would be..

1) choosing ways where all 5 are of one kind and subtracting from all ways... 
 total ways-  10 C 5
 all one of a kind- 6 C 5
 total selections= 10!5!5!−6!5!1!=246
 D

Very easy thing to miss on, but hope it helps someone

To add to this, selecting only paperbacks is not possible as it is only 4 books and if we select 4 paperbacks then 1 would automatically be hardback which is not what we are looking to subtract.

luisdicampo · Answer

Deconstructing the Question

We have  4 paperbacks  and  6 hardbacks , and we choose 5 books.

We want selections that contain at least one of each type. Use the complement: subtract invalid cases from total.

Step-by-step

Total ways to choose 5 books from 10:

\binom{10}{5} = 252

Now subtract the invalid cases.

All paperbacks:

\binom{4}{5} = 0

All hardbacks:

\binom{6}{5} = 6

So valid selections:

252 - 6 = 246

Answer D

There are 10 books on a shelf, of which 4 are paperbacks and 6 are har

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