The only contents of a container are 10 disks that are each numbered

Question

The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7 ?

A) 1/7

B) 3/14

C) 2/7

D) 1/2

E) 15/28

(A) 1/7
(B) 3/14
(C) 2/7
(D) 1/2
(E) 15/28

KarishmaB · Accepted Answer

Range = 7 = Largest Number - Smallest Number

There are 3 pairs of numbers that can give us this range: (8, 1), (9, 2) and (10, 3)
For each pair, there are 6 numbers between the limits out of which 2 need to be picked up e.g. we need to pick 2 of (7, 6, 5, 4, 3, 2) if our pair is (8, 1). This will give us 4 numbers.
Hence no of cases = 3 * 6C2

Total no of ways of picking 4 numbers out of 10 = 10C4

Req Probability = 3*6C2 / 10C4 = 3/14

IanStewart · Answer

It's neither a combinations nor a permutations problem; it's a probability problem.

In general, in a probability question where you are selecting things without replacement, you can pretend order matters, or pretend it doesn't, and you'll get the same answer either way. This is easier to see using a simpler question: Say you have a bag containing 7 blue marbles and 5 red marbles, and you want to know the probability you will get 2 blue marbles if you randomly pick two different marbles from the bag. You can imagine putting both hands in the bag, and grabbing two different marbles. Whether you lift your hands out one at a time (so there is a first selection and a second, and order matters) or take both hands out at once (so order does not matter), the probability you'll get two blue marbles must be the same - you're taking out the same two marbles either way. So you can answer this probability question in three ways:

- you can just use the probability rules, which is what I would do, though when we do this we're implicitly assuming we're picking in order: the probability the first marble is blue is 7/12, and then the probability the second is also blue is 6/11, so the probability both are blue is (7/12)(6/11) = 7/22

- we can use the definition of probability (which is what the solutions in this thread are using) when events are equally likely: the probability X happens = (number of ways X can happen) / (total number of things that can happen). When we use this definition for a problem like this, we can pretend order matters, or we can pretend order doesn't matter. As long as we're consistent, i.e. as long as we count the numerator and denominator under the same assumption, we'll get the right answer.

- So if we pretend order matters, we'd say "there's 7 ways for the first marble to be blue, and 6 ways for the second to be blue, and thus 7*6 ways for the first two selections to be blue, and there are 12*11 ways in total to pick two marbles in order, so the answer is 7*6/12*11 = 7/22

- Or we can pretend order does not matter. Then there are 7C2 = 7*6/2! ways to pick a set of two blue marbles, and 12C2 = 12*11/2! ways to pick a set of any two marbles, and the probability both are blue is 7C2/12C2 = (7*6/2) / (12*11/2) = 7/22

The same principle applies to the question in this thread, though I'd find this particular problem easier to solve by assuming order does not matter. Then we'd have 10C4 ways to pick four numbers. To get a range of 7, our smallest and largest numbers could be '1' and '8', and then we'd have 6C2 ways to pick our other two numbers (from the set of numbers between 2 and 7 inclusive). But our smallest and largest numbers can be 2 and 9, or can be 3 and 10, so we have 3*6C2 ways to get a range of seven, and the answer is 3*6C2/10C4 = (3)(3)(5) / (10)(3)(7) = 3/14

But you could pretend order matters instead. Then there's 10*9*8*7 ways to pick four numbers. There are going to be many ways to get a range of exactly 7. We might pick:

1, 8, X, Y

where X and Y are between 2 and 7 inclusive. Since we have 6 choices for X and 5 choices for Y, there's 30 ways to pick numbers in this precise order. But we could also pick our numbers in these orders:

1X8Y
1XY8
X18Y
X1Y8
XY18

so there are six different orders we could have of 1, 8, X and Y, for 6*30 = 180 orders, and we need to double that, because the '8' could be selected before the '1' (e.g. we could have 81XY instead of 18XY). So there are 360 ways to pick, in order, four numbers with a range of seven when the smallest selection equals 1. But the smallest selection can be 2, and can be 3, so there are three equally likely ways in total to get a range of seven, and there will be 3*360 ways to pick four numbers in order that give a range of seven. So the answer is

(3)(360)/(10)(9)(8)(7) = (3)(36)/(9)(8)(7) = 3/14

That's a correct way to solve the problem, but it's long, which is why the solutions above pretended order did not matter instead. On easier problems of this type, it's almost always easiest to assume order matters, but on harder problems, like this one, it can be easier to assume it doesn't.

Raj30 · Answer

It is mentioned in the question that disks are picked without replacement. How can we use nCk here?
In the numerator instead of 6C2 it should be 6x5 (Choose 1 fom 6 then 1 from remaining 5,ie 6C1 X 5C1)
Similarly, in the denominator instead of 10C4 it should be 10x9x8x7
The answer comes out to be 3/28 by this approach
Where am i going wrong?

allornothing21 · Answer

Posted from my mobile device

In the numerator, it has to be 6c2 because you are not concerned about the numbers picked or their order for that matter. You are picking ranges of (1,8) (2,9) and (3,10). In all these three cases you have to make sure that the range is being satisfied and you can pick remaining two disks from the remaining six numbers that can come between 1 and 8 and so on. You are not concerned with the order here. See had it been a case of arranging the items, your approach would have been right. But here we are not concerned about that but the selection.

So in this case, the numerator would be 6c2 for one case (1,8) and for the rest of the cases it would be

3*6c2

And for denominator, again, since you are not concerned about the order, you are just selecting 4 numbers out of 10 which is

10c4

And the answer would be 3/14

Hope it makes sense.

Kavish1710 · Answer

Can someone please elaborate on the solution. How come we are only considering 6C2. How are we calculating the probability of the extreme two disks.

Posted from my mobile device

aayush.works · Answer

Before we start on this question, lets keep in mind that range is derived in a "ordered set". Thus the arrangement of numbers cannot be shuffled (which means 3, 7,8,10 can be one possibility 
but 3,8,7,10 cannot be correct according to the constraints given)

Step1:-
First let us draw the cases possible wherein the range of 7 is possible:-

1) 3 _ _ 10 
(1,2 will not be permitted)

2) 2 _ _ 9
(1,10 will not be permitted)

3) 1 _ _ 8
(9,10 will not be permitted)

No other possibilities exist.

Step2 :-
For each option we have 6C2 choices.

(A)Total desirable outcomes : 15*3 = 45
(B)Total possible outcomes : 10C4
Probability(A/B) : =3/14

Archit3110 · Answer

total possible pairs which have range of 7 ; (3,10,) ; (2,9); (1,8) viz 3 
and total fair chances for drawing such nos; 4/10*3/9*2/8*1/7 ; 1/210 
since we know 2 no of each pair (3,10,) ; (2,9); (1,8) ; the other 2 no can be determined by ; 6c2 ; 15
total possibilities ; 15*3/210 ; 3/14
IMO B

NoHalfMeasures · Answer

you might be upto something here Raj30!

You 're right in your approach. Here is my complete solution-

There will be 3 types of favourable outcomes:
1) 1, 8 and other 6 number from 2 to 7
2) 2, 9 and other 6 number from 3 to 8
3) 3, 10 and other 6 from 4 to 9

Let's consider case 1) -
No of ways of choosing 8, 1 = 4*3 (4p2)
No of ways of choosing 2 numbers from other 6 numbers = 6*5 (6p2)

No of ways case1 can happen = 4*3*6*5

No of ways 3 cases can happen = 4*3*6*5 *3

Total possible outcomes = 10p4 = 10*9*8*7

Probability = 4*3*6*5 *3 / 10*9*8*7 = 3/14

Hope it helps

vaibhav1221 · Answer

Minimum possible range = 3, maximum possible range = 9
A range of 3 is possible in 7 cases (from 1-4 to 7-10)
Similarly
A range of 4 is possible in 6 cases
A range of 5 is possible in 5 cases
A range of 6 is possible in 4 cases
A range of 7 is possible in 3 cases
A range of 8 is possible in 2 cases
A range of 9 is possible in 1 case

Total cases = 1+2+3+4+5+6+7 = 28
Favorable cases = 3
Probability = 3/28
Now, the smallest number in the range can be chosen first or the largest one can be chosen first. So, there would be a case where the smallest is chosen first (probability = 3/28) and there would be a case where the largest is chosen first (probability = 3/28)
(We are not concerned about the numbers in the middle in this solution. We are just concerned about the smallest and largest number.)
3/28+3/28 = 3/14

missionmba2025 · Answer

KarishmaB  - here is how I approached the question. Can you please let me know what I am missing in the first approach-

_ _ _ _

First Approach : Using Probability

Step 1: Fill the last digit - To obtain a range of 7, we must have either 10, 9 or 8 in the last digit. The probability of getting a 10, 9 or 8 in the last digit = 3/10
Step 2: Fill the first digit - The first digit can be filled only in 1 way. The probability of getting that digit = 1/9
Step 3: The middle two digits can be filled in 6 ways and 5 ways respectively. Their probability = 6/8 * 5/7
Step 4: The four numbers can be arranged in 4! ways

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

Not sure what I am missing here.

Second Approach : Using Filling Space Method

Step 1: Fill the last digit - To obtain a range of 7, we must have either 10, 9 or 8 in the last digit. Hence, the last digit can be filled in 3 ways.
Step 2: Fill the first digit - The first digit can be filled only in 1 way.
Step 3: The middle two digits can be filled in 6C2 = 6 * 5/2 = 3 * 5
Step 4: The four numbers can be arranged in 4! ways. Number of ways to perform the selection and arrange the selected numbers = 3 * 3 * 5 * 4! 
Step 5: Total number of possible selections - Four numbers can be selected in 10C4 ways and can be arranged 4! ways. Number of ways to select and arrange 4 numbers = 10C4 * 4! = 10 * 9 * 8 * 7

Total Probability = 3 * 3 * 5 * 4! / 10 * 9 * 8 * 7 = 3/14

I am unable to figure out what I am doing wrong in the first method. Could you please help me.

KarishmaB · Answer

Here is something for you to think about in your first approach:

Step 1: Fill the last digit - To obtain a range of 7, we must have either 10, 9 or 8 in the last digit. The probability of getting a 10, 9 or 8 in the last digit = 3/10

Say you selected 10.

Step 2: Fill the first digit - The first digit can be filled only in 1 way. The probability of getting that digit = 1/9

You must select 3.

Step 3: The middle two digits can be filled in 6 ways and 5 ways respectively. Their probability = 6/8 * 5/7

Now you selected 5 (since you can select any 1 out of 4 to 9) and then you selected 7 (now you select any one out of 4 to 9 except 5.) 
Another case would be in which you selected 7 (since you can select any 1 out of 4 to 9) and then selected 5 (now you select any one out of 4 to 9 except 7.)

Step 4: The four numbers can be arranged in 4! ways

You are arranging 10, 3 , 5 and 7 in 4! ways. But you have already counted 57 and 7 5 as two different cases. So you must divide by 2!

missionmba2025 · Answer

Thank you so much  KarishmaB  for your reply. You're the best !

isteducimus · Answer

KarishmaB   - here is how I approached the question. Can you please let me know what I am missing -

Step 1: 3 possible scenarios - 10,_,_,3 and 9,_,_,2 and 8,_,_,1. Can not select 2 digits out of the rest for each pair. So 1 digit can be filled in 6 ways, another in 5, rest 2 digits are fixed for each pair. So we have 6*5*1*1 - possible digits selection for each pair 
 Step 2: The four numbers can be arranged in 4! ways. Number of ways to perform the selection and arrange the selected numbers = 3*6*5*4! 
 Step 3: Total number of possible selections - 10*9*8*7

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

I am unable to figure out what I am doing wrong. Could you please help me.

egmat · Answer

isteducimus

Looking at your approach, you've done excellent work identifying the three scenarios correctly! Your mistake lies in how you're counting the arrangements, which is causing you to double-count. Let me help you see where the issue is.

Your Error Diagnosis:

You correctly identified that for range  = 7 , we need:
-  (10, -, -, 3)  or  (9, -, -, 2)  or  (8, -, -, 1)

However, when you calculated  6 	imes 5 	imes 4!  for arrangements, you're actually counting each selection multiple times. Here's why:

The Issue with Your Counting:

When you select the middle numbers as  6 	imes 5 , you're already considering ORDER (first selecting one number, then another). But then you multiply by  4!  to arrange all four numbers, which counts the ordering AGAIN.

For example, if you select numbers  10, 7, 5, 3 :

- Your  6 	imes 5  counts selecting  7  first then  5  as different from selecting  5  first then  7 
- But then  4!  also rearranges these same numbers
- This creates duplicate counting!

The Correct Approach:

Method 1: Using Combinations (Simpler)

For each scenario (e.g., max  = 10 , min  = 3 ):
- We must include  10  and  3 
- Choose  2  numbers from  \{4, 5, 6, 7, 8, 9\} 
- Number of ways  = C(6,2) = 15

Total favorable outcomes  = 3 	imes 15 = 45 
Total possible selections  = C(10,4) = 210

Probability  = \frac{45}{210} = \frac{3}{14}

Method 2: Fixing Your Ordered Approach

If you want to use ordered counting throughout:
- Favorable outcomes:  3 	imes 6 	imes 5 	imes 2!  (not  4! , because two positions are fixed)
- But wait! This gives  3 	imes 6 	imes 5 	imes 2 = 180

Actually, for ordered selection:
- Each scenario: Select and arrange  4  numbers where  2  are fixed
- This is  P(6,2) 	imes 4!/2!  ways to include the middle numbers and arrange everything
-  = 30 	imes 12 = 360  per scenario
- Total:  3 	imes 360 = 1080

Total ordered selections:  P(10,4) = 10 	imes 9 	imes 8 	imes 7 = 5040

Probability  = \frac{1080}{5040} = \frac{3}{14}

Why You Got  3/7 :

Your calculation  \frac{3 	imes 6 	imes 5 	imes 4!}{10 	imes 9 	imes 8 	imes 7} = \frac{3 	imes 6 	imes 5 	imes 24}{5040} = \frac{2160}{5040} = \frac{3}{7}

This is exactly double the correct answer because you counted each arrangement twice through the mixed ordering approach!

You can practice similar counting and probability problems  here  - focus on combinatorics questions to reinforce consistent counting approaches.

Key Takeaway: 
When counting, be consistent: either count  ordered  selections throughout OR count  unordered  selections throughout. Mixing the two methods leads to overcounting or undercounting.

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