A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6

Question

A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3

A. 1/60 
B. 1/30
C. 1/20
D. 1/6
E. 1/3

M37-38

stne · Accepted Answer

Hi  riagarg07 ,

Let me try and explain.

From 6 cards if we were to select 3 cards we can do that in 6C3 or 20 ways.  NOTE  this will give us the total number of 3 card sets .e.g. 645, 321 are just two of such sets among the 20 sets.
 
Now in each of these sets only  ONE ARRANGEMENT  will be in ascending order.
e.g. 645 can be arranged in total 3! or 6 ways i.e.

456
654
645
546
465
565

Out of these only one set (456) is in ascending order.

Hence  EVERY   3 card set will have only one way in which that set can be arranged in ascending order.
Now we have a total of 6C3 or 20 sets and in each set there is only one way in which that set can be in ascending order.

Hence total number of ways in which 3 cards can be chosen from 6 cards such that the cards are in ascending order is 
6C3 *1

hope it's clear.

Bunuel · Answer

Official Solution:

A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3

A.  \frac{1}{60} 
B.  \frac{1}{30} 
C.  \frac{1}{20} 
D.  \frac{1}{6} 
E.  \frac{1}{3}

Any three cards can be drawn in  3! = 6  ways, and only one sequence from these 6 will be in ascending order. So, the probability is  \frac{1}{6} .
 
For example, 2, 5, and 6 can be drawn in 6 ways ({firs card, second card, third card}): {2, 5, 6}, {2, 6, 5}, {5, 2, 6}, {5, 6, 2}, {6, 2, 5} and {6, 5, 2}. Only one from these 6 is in ascending order: {2, 5, 6}.

Answer: D

tbhauer · Answer

In order for the 3 cards to be in sequence, the 1st card drawn must be a 1, 2, 3 or 4, so the odds of drawing one of these cards is 4/6. Without replacement, the next card must be in sequence, so the odds for drawing the next card in sequence is 1/5. Without the replacement, the same rules apply for the next card, so the odds of drawing the next card in sequence is 1/4. Total odds = 4/6 * 1/5 * 1/4 = 4/120 = 1/30
Ans= B

ShankSouljaBoi · Answer

Alternative approach :

Let the three cards be a , b and c

So the question becomes 
is a<b<c ?

Now, a can have any three of the operations < = > with b and b can have any two operations with c
So total options = 3*2
Favourable: a<b<c ... Only one outcome.
Probablity = 1/6
D

Posted from my mobile device

nick1816 · Answer

Total number of possible arrangements of 3 cards = 6C3 *3!

Total number of ways in which 3 card are in the increasing order = 6C3 * 1

Probability = 1/3! = 1/6

ScottTargetTestPrep · Answer

The number of ways to select 3 cards form 6 where order matters is 6P3 = 6 x 5 x 4 = 120. Let’s use a table to list the number of ways in which 3 numbers are in increasing order if the first 2 numbers (in increasing order) are drawn:

First two numbers Number of ways
 1, 2 4
 1, 3 3
 1, 4 2
 1, 5 1
 2, 3 3
 2, 4 2
 2, 5 1
 3, 4 2 
 3, 5 1
 4, 5 1

We see that there are 10 + 6 + 3 + 1 = 20 ways in which 3 numbers can be drawn without replacement such that they are in increasing order (notice that each addend is the number of ways where the smallest of the 3 numbers is 1, 2, 3 and 4, respectively). Therefore, the probability is 20/120 = 1/6.

Answer: D

CAMANISHPARMAR · Answer

You can choose 3 cards from 6 cards and arrange them in 6P3 ways; i.e. 120 ways. (Permutations)

You can choose 3 cards from 6 cards and arrange them in one way, i.e. 6C3; i.e. 20 ways; that is any unique combination of three numbers will effectively be in increasing order.

Ans is 20/120 = 1/6 (Option D)

vinny12 · Answer

Hi Nick1816,
Can you please explain how have you counted the number in the increasing order as 6C3. Though I got the same result by counting the possible arrangements but I want to understand the thought process. Please explain.

Vinay

nick1816 · Answer

Hey
Suppose you choose 1, 2 and 3.

Total possible arrangements of 1, 2 and 3 are 3!= 6.

But there is only 1 possible arrangement out of those 6 arrangements in which you can put these numbers in increasing order, i.e (1,2,3), and this is true for any combination of 3 distinct numbers.

Hence, Total number of ways in which 3 card are in the increasing order =   *  = 6C3 * 1

If you still have any doubt, you can ask.

Hrley · Answer

I have an interesting way of approaching this question.

We're asked about the probability that the 3 numbers we choose are in increasing order,
so this means our denominator does include the order of those numbers while our numerator should not.

So, we can interpret the answer as 6C3/6P3 = 1/6.

I initially approached this question by automatically writing the denominator as 6C3 which made it difficult to understand why it's 1/6 when 20 isn't divisible by 6...

A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6

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