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

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Math Expert
Joined: 02 Sep 2009
Posts: 65187
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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22 Apr 2020, 10:37
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Difficulty:

45% (medium)

Question Stats:

67% (01:58) correct 33% (01:52) wrong based on 152 sessions

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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 consecutive integers in increasing order?

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

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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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22 Apr 2020, 12:16
1
1
Total number of possible arrangements of 3 cards = 6*5*4 = 120

Number of ways to select consecutive integers cards in increasing order = 4 [{1,2,3}, {2,3,4}, {3,4,5}, {4,5,6}]

Probability = 4/120 = 1/30

Bunuel wrote:
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 consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3
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Posts: 6438
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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23 Apr 2020, 03:53
Bunuel wrote:
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 consecutive integers in increasing order?

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

total ways to select 3 cards from 6 ; 6*5*4 ; 120 ways
and possible set of consective cards ; (1,2,3) ; ( 2,3,4); ( 3,4,5); ( 4,5,6) ; 4
so P = 4/120 ; 1/30
OPTION B
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Joined: 23 Feb 2020
Posts: 46
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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21 May 2020, 21:12
1
Bunuel wrote:
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 consecutive integers in increasing order?

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

Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device
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WE: Education (Education)
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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22 May 2020, 01:57
Mck2023 wrote:
Bunuel wrote:
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 consecutive integers in increasing order?

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

Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device

Mck2023

Understand that permulation formula includes selection as well as arrangements

Total Outcomes = Permutation of 3 out of 6 digits = 6P3 = 6! / (6-3)! = 120

Favorable outcomes = {123} or {234} or {345} or {456} = 4 outcomes

Probability = 4/120 = 1/30

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Joined: 23 Feb 2020
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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22 May 2020, 19:31
GMATinsight wrote:
Mck2023 wrote:
Bunuel wrote:
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 consecutive integers in increasing order?

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

Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device

Mck2023

Understand that permulation formula includes selection as well as arrangements

Total Outcomes = Permutation of 3 out of 6 digits = 6P3 = 6! / (6-3)! = 120

Favorable outcomes = {123} or {234} or {345} or {456} = 4 outcomes

Probability = 4/120 = 1/30

Hi GMATinsight

I am still not clear what in the question make us use permutation formula but nit the combination one?

Thank you!
GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4345
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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22 May 2020, 20:57
Mck2023 wrote:
GMATinsight wrote:
Mck2023 wrote:
Bunuel wrote:
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 consecutive integers in increasing order?

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

Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device

Mck2023

Understand that permulation formula includes selection as well as arrangements

Total Outcomes = Permutation of 3 out of 6 digits = 6P3 = 6! / (6-3)! = 120

Favorable outcomes = {123} or {234} or {345} or {456} = 4 outcomes

Probability = 4/120 = 1/30

Hi GMATinsight

I am still not clear what in the question make us use permutation formula but nit the combination one?

Thank you!

Mck2023

Use Combination when you need to do ONLY SELECTION

Use Permutaton when you need to do SELECTION + ARRANGEMENT
_________________
Prepare with PERFECTION to claim Q≥50 and V≥40 !!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Most affordable l Comprehensive l 2000+ Qn ALL with Video explanations l LINK: Courses and Pricing
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
FREE GMAT Resource: 22 FREE (FULL LENGTH) GMAT CATs LINKS l NEW OG QUANT 50 Qn+VIDEO Sol.
Manager
Joined: 16 Jan 2020
Posts: 55
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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30 May 2020, 17:45
To select consecutive integers in increasing order, the possible sets are as follows:

{1, 2, 3} OR
{2, 3, 4} OR
{3, 4, 5} OR
{4, 5, 6}

Next, we note that the probability of selecting any of the 4 sets above is the same. So we can simply multiply the probability of 1 set by 4.

Selecting the 1st card is $$\frac{1}{6}$$
Selecting the 2nd card is $$\frac{1}{5}$$
Selecting the 3rd card is $$\frac{1}{4}$$

P = 4 x $$\frac{1}{6}\frac{1}{5}\frac{1}{4}$$
P = $$\frac{1}{30}$$

Hope it helps.
Intern
Joined: 11 Nov 2018
Posts: 13
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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11 Jun 2020, 22:27
Cellchat wrote:
To select consecutive integers in increasing order, the possible sets are as follows:

{1, 2, 3} OR
{2, 3, 4} OR
{3, 4, 5} OR
{4, 5, 6}

Next, we note that the probability of selecting any of the 4 sets above is the same. So we can simply multiply the probability of 1 set by 4.

Selecting the 1st card is $$\frac{1}{6}$$
Selecting the 2nd card is $$\frac{1}{5}$$
Selecting the 3rd card is $$\frac{1}{4}$$

P = 4 x $$\frac{1}{6}\frac{1}{5}\frac{1}{4}$$
P = $$\frac{1}{30}$$

Hope it helps.

Here then you have assumed that the order you have picked up the cards is also consecutive

Which is not the case mentioned in the question. The question says that the cards picked up line up in order afterwards.
Manager
Joined: 12 Mar 2019
Posts: 164
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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12 Jun 2020, 11:43
Mck2023 wrote:
GMATinsight wrote:
Mck2023 wrote:
Bunuel wrote:
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 consecutive integers in increasing order?

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

Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device

Mck2023

Understand that permulation formula includes selection as well as arrangements

Total Outcomes = Permutation of 3 out of 6 digits = 6P3 = 6! / (6-3)! = 120

Favorable outcomes = {123} or {234} or {345} or {456} = 4 outcomes

Probability = 4/120 = 1/30

Hi GMATinsight

I am still not clear what in the question make us use permutation formula but nit the combination one?

Thank you!

Permutation is used, when ORDER MATTERS.
In this question, we need to use PERMUTATION, because we need probability of cards to be in ascending order, after we draw first card.
e.g if first card drawn is 3, 2nd card should be 4 - not any other, So order matters. I hope you understood.
Total cases formed are thus 4 / 6p3 : 4/120 = 1/30

Combination is used when order doesn't matter and you have to just find total combination or ways of arrangement of anything.

Kudos if it helped !!!
Intern
Joined: 30 May 2019
Posts: 7
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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14 Jun 2020, 04:19
1st number from the has to be amongst {1,2,3,4}
Therefore probability of 1st number is 4/6.

2nd number has to be only one of the balance five numbers I.e. next consecutive number. Therefore probability of second number is 1/5

Likewise probability of third number will be 1/4.

So,final answer I.e. probability of three consecutive numbers among given six numbers in increasing order will be 4/6 * 1/5 * 1/4 = 1/30

Posted from my mobile device
Manager
Joined: 16 Jan 2020
Posts: 55
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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14 Jun 2020, 13:52
PrajwalGupta wrote:
Cellchat wrote:
To select consecutive integers in increasing order, the possible sets are as follows:

{1, 2, 3} OR
{2, 3, 4} OR
{3, 4, 5} OR
{4, 5, 6}

Next, we note that the probability of selecting any of the 4 sets above is the same. So we can simply multiply the probability of 1 set by 4.

Selecting the 1st card is $$\frac{1}{6}$$
Selecting the 2nd card is $$\frac{1}{5}$$
Selecting the 3rd card is $$\frac{1}{4}$$

P = 4 x $$\frac{1}{6}\frac{1}{5}\frac{1}{4}$$
P = $$\frac{1}{30}$$

Hope it helps.

Here then you have assumed that the order you have picked up the cards is also consecutive

Which is not the case mentioned in the question. The question says that the cards picked up line up in order afterwards.

The question states "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 consecutive integers in increasing order?"

You must pick the cards up in increasing order while at the same time pick consecutively.

There's 4 different ways you can do this, each of which has a probability of $$\frac{1}{120}$$.
Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6   [#permalink] 14 Jun 2020, 13:52