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In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we

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In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we randomly choose 4 of the discs, what is the probability that the range of the numbers chosen is 7?

(A) 1/14
(B) 1/7
(C) 3/14
(D) 9/28
(E) 4/7

[Reveal] Spoiler:
Correct answer is (C)

I can do this as a permutations problem, but the given explanation does it as a combinations problem,
and that I'm not quite following.

Permutations: Calculate total first (probability = desired outcomes / total outcomes); choosing 4 discs from 10 is simple: 10 x 9 x 8 x 7 possiblities = 5040.
Now to figure out the 4 discs.
There are 3 distinct possibilities (discs "1" & "8", "2" & "9", or "3" & "10") if the range = 7. So we will analyze one of these, then we'll need to multiply the result by 3. Let's take discs "1" & "8".
There are 4 possible places for disc "1" (choice 1 or 2 or 3 or 4) and then 3 places for disc "8". So 4 x 3 = 12 different arrangements possible for the fixed numbers.
For each of the 12 arrangements, we still have to choose the other 2 discs, which must be 2 of the 6 remaining numbers in between (discs "2","3","4","5","6","7").
For the first of these 2 choices, there are 6 possiblities, then 5 possibilities for the other.
SO: desired outcomes are 4 x 3 x 6 x 5 (all the possibilities if choosing discs 1 & 8 & two others) x 3 (if choosing 2 & 9 or 3 & 10 instead of 1 & 8).
This yields (6 x 5 x 4 x 3 x 3) / (10 x 9 x 8 x 7) = 3 / 14, which is the answer.

Combinations: calculate total; choosing 4 from 10 (order does not matter now) = 10C4 = 210.
There are still 3 distinct possiblities (discs "1" & "8", "2" & "9", or "3" & "10") to consider; evaluate one and then x 3.
Let's choose discs "1" & "8". Since there will be a total of 4 discs chosen, 2 of them must still be chosen & will have to be chosen from the 6 discs numbered in between 1 & 8 (discs "2","3","4","5","6","7"). That's choosing 2 from 6, or 6C2 = 15.

At this point, the official explanation says "Thus, the total number of possibilities when choosing 4 discs that include discs 1 & 8 and two numbers in between is 15."
This is the part I do not understand. How did we get from probability for two discs to probability for all 4?

Finishing up, desired outcomes are 15 x 3 = 45, and probability in toto is 45 / 210 = 3 / 14.

Clearly combinations is simpler & faster; I just need to comprehend it fully. Help!
[Reveal] Spoiler: OA

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Re: In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we [#permalink]

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axs wrote:
In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we randomly choose 4 of the discs, what is the probability that the range of the numbers chosen is 7?
(A) 1 / 14
(B) 1 / 7
(C) 3 / 14
(D) 9 / 28
(E) 4 / 7


Welcome to GMAT Club. Below is combinations approach to this question, hope it helps to clear your doubts

In order the range to be 7, the greatest and smallest numbers out of 4 we pick should be: (10,3), (9,2), or (8,1). Notice that the other two numbers can be any within this range.

Let's consider one specific case (10,3). In how many ways we can pick 10, 3, and ANY 2 other numbers between them? \(C^1_1*C^2_6=15\), where \(C^1_1=1\) is one way to pick 3 and 10, and \(C^2_6\) is # of ways we can pick 2 other different numbers out of 6 between 3 and 10 (out of 4, 5, 6, 7, 8, and 9).

The same for other two cases, thus # of favorable outcomes is 3*15=45.

Total # of outcomes \(C^4_{10}=210\), where \(C^4_{10}\) is # of ways we can pick 4 different numbers out of 10.

P=Favorable/Total=45/210=3/14

Answer: C.

Hope it's clear.

P.S. Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.
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New post 01 Feb 2012, 06:13
Thanks for a recapitulation of the combinations method for solving this problem.
Unfortunately, I still do not understand the critical step.

Yes, choosing any 2 from 6 (6C2) = 15.
But then you say "1C1 (= 1) is one way to pick 3 and 10". Huh?
It's not making sense to me.
If I say that I want to know the probability of choosing disc "3", I'm choosing 1 disk from among 10.
How is that a probability of 1?

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Re: In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we [#permalink]

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New post 01 Feb 2012, 06:26
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axs wrote:
Thanks for a recapitulation of the combinations method for solving this problem.
Unfortunately, I still do not understand the critical step.

Yes, choosing any 2 from 6 (6C2) = 15.
But then you say "1C1 (= 1) is one way to pick 3 and 10". Huh?
It's not making sense to me.
If I say that I want to know the probability of choosing disc "3", I'm choosing 1 disk from among 10.
How is that a probability of 1?


Forget about the probability for a moment. The question is in how many ways we can choose 3, 10, and ANY 6 number between them? In how many ways you can choose one number 3 out of 10? In one way. In how many ways you can choose one number 10 out 10? In one way. How else?

Consider simpler example: in how many ways we can choose an orange out of 1 orange, 1 apple and 1 banana? In one way: to pick this orange. The probability of picking an orange would be 1/3, so 1 the ways we can pick an orange (favorable outcome) and 3 is total # of outcomes.

The same above: we can choose (3, 10) in one way: as we can pick 3 in one way and we can pick 10 in one way.

Hope it's clear.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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axs wrote:
Thanks for a recapitulation of the combinations method for solving this problem.
Unfortunately, I still do not understand the critical step.

Yes, choosing any 2 from 6 (6C2) = 15.
But then you say "1C1 (= 1) is one way to pick 3 and 10". Huh?
It's not making sense to me.
If I say that I want to know the probability of choosing disc "3", I'm choosing 1 disk from among 10.
How is that a probability of 1?


Let me add something here:

The terms Combinations and Probability are related but different.

Probability (A) = No of suitable outcomes where A happens/Total no of outcomes

We use combinations to get "No of suitable outcomes where A happens" and "Total no of outcomes" and then calculate the required probability. So for the time being, just forget that you have to find some probability.
Focus on two things: "No of suitable outcomes where A happens" and "Total no of outcomes"

It is easy to find total number of outcomes. 10C4. e.g. (2, 3, 4, 6), (4, 1, 9, 5) etc etc etc

In how many combinations is range 7? (This is the "No of suitable outcomes where A happens")
(3, 10, a, b), (2, 9, c, d) and (1, 8, e, f)
a and b are numbers between 3 and 10 (so that range doesn't exceed 7)
c and d are numbers between 2 and 9
e and f are numbers between 1 and 8

In how many ways can you choose a and b? 6C2 (Any 2 out of 6 numbers between 3 and 10). So how many 4 number combinations are there which look like this: (3, 10, 4, 6), (3, 10, 7, 5) etc?
I hope you agree it is 6C2.
Similarly, you get 6C2 combinations of the type (2, 9, c, d) and 6C2 of the type (1, 8, e, f).

No of suitable combinations = 6C2 + 6C2 + 6C2

Now, the required probability = No of suitable combinations where A happens/Total no of outcomes
= (3*6C2)/10C4
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Re: In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we [#permalink]

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New post 22 Jan 2015, 09:58
VeritasPrepKarishma wrote:
axs wrote:
Thanks for a recapitulation of the combinations method for solving this problem.
Unfortunately, I still do not understand the critical step.

Yes, choosing any 2 from 6 (6C2) = 15.
But then you say "1C1 (= 1) is one way to pick 3 and 10". Huh?
It's not making sense to me.
If I say that I want to know the probability of choosing disc "3", I'm choosing 1 disk from among 10.
How is that a probability of 1?


Let me add something here:

The terms Combinations and Probability are related but different.

Probability (A) = No of suitable outcomes where A happens/Total no of outcomes

We use combinations to get "No of suitable outcomes where A happens" and "Total no of outcomes" and then calculate the required probability. So for the time being, just forget that you have to find some probability.
Focus on two things: "No of suitable outcomes where A happens" and "Total no of outcomes"

It is easy to find total number of outcomes. 10C4. e.g. (2, 3, 4, 6), (4, 1, 9, 5) etc etc etc

In how many combinations is range 7? (This is the "No of suitable outcomes where A happens")
(3, 10, a, b), (2, 9, c, d) and (1, 8, e, f)
a and b are numbers between 3 and 10 (so that range doesn't exceed 7)
c and d are numbers between 2 and 9
e and f are numbers between 1 and 8

In how many ways can you choose a and b? 6C2 (Any 2 out of 6 numbers between 3 and 10). So how many 4 number combinations are there which look like this: (3, 10, 4, 6), (3, 10, 7, 5) etc?
I hope you agree it is 6C2.
Similarly, you get 6C2 combinations of the type (2, 9, c, d) and 6C2 of the type (1, 8, e, f).

No of suitable combinations = 6C2 + 6C2 + 6C2

Now, the required probability = No of suitable combinations where A happens/Total no of outcomes
= (3*6C2)/10C4


VeritasPrepKarishma

Hi, I would really appreciate it if you could please point out the flaw in my approach:

1/10 * 1/9 * 6/8 * 5/7 * 4! * 3 = 3/7

I understand that you have a 2! in the denominator on account your combination approach but how do we address that in this method. ?
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Re: In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we [#permalink]

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SahilG wrote:
Hi, I would really appreciate it if you could please point out the flaw in my approach:

1/10 * 1/9 * 6/8 * 5/7 * 4! * 3 = 3/7

I understand that you have a 2! in the denominator on account your combination approach but how do we address that in this method. ?


There is a reason Bunuel and I have used combinations approach - it is far less complicated to use combinations in cases where you pick multiple objects in a go.

The way I see your approach is this:

1/10 (choose first disc, say 1) * 1/9 (choose corresponding second disc, say 8) * 6/8 (choose third disc from in between numbers, say 5) * 5/7 (choose fourth disc from leftover in between numbers, say 6) * 4! (arrange the 4 numbers 1,8,5,6) * 3 (choose the first disc in 3 ways 1, 2 or 3)

The problem here is 6/8 * 5/7. It already includes the re-arrangement of the two numbers chosen.

Say, we select 1 and 8 as first two discs.
Leftover discs: 2, 3, 4, 5, 6, 7, 9, 10
The third disc is selected => 6/8. Say we select number 5.
The fourth disc is selected => 5/7. Say we select number 6.

Another case will be
The third disc is selected => 6/8. Say we select number 6.
The fourth disc is selected => 5/7. Say we select number 5.

So we have already accounted for arrangement of these 2 discs. But later we have multiplied by 4! to account for arrangement of all 4 numbers. Hence, when we have 6/8*5/7, we need to divide it by 2 to eliminate the repeat picks.

Hope it makes sense now.
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Re: In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we [#permalink]

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New post 13 Apr 2016, 09:22
I got 3 possibilities for range 7. But the total possibilities, I get 28. How is that 14?

Could someone please help me with where I'm going wrong?

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Re: In a bag, there are 10 discs, numbered 1 to 10 (no duplicates). If we [#permalink]

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dina98 wrote:
I got 3 possibilities for range 7. But the total possibilities, I get 28. How is that 14?

Could someone please help me with where I'm going wrong?


The possibilities for range 7 are not 3; they are 45. And the total possibilities are 210.
3/14 is 45/210 in lowest terms. Check the solutions discussed above.
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