How Many Unique 7-Digit Numbers Can Be Formed Using Specific Repeated Digits?

In summary: I think it makes more sense in the way it is used in post 3, but then it is kind of trivial (otherwise it would not make sense to list some numbers twice). @Mohd Taqi: Can you please clarify what "no repetition" means?In summary, the digits 1, 1, 2, 2, 4, 4 and 5 can form 7-digit numbers if repetition is not allowed. If one of these numbers is chosen at random, find the probability that it is (a) greater than 4,000,000(b) even number and greater than 4,000,000.The digits 1, 1, 2, 2, 4, 4 and 5 can form 7-digit numbers if
  • #1
Mohd Taqi
1
0
How many 7-digit numbers can be formed from the digits 1, 1, 2, 2, 4, 4 and 5 if repetition is not allowed. If one of these numbers is chosen at random, find the probability that it is(a) greater than 4,000,000(b) even number and greater than 4,000,000

Question b is so confusing . Help me
 
Physics news on Phys.org
  • #2
How did you solve question (a)? A modification of that should allow to answer (b).

How can you tell if a number is even? For example, is 46354235366534213232 even, and how do you know that without using a calculator?

"Repetitions not allowed" means numbers like 1122445 are not allowed, but 1241245 is okay because ther are other digits between the two "1", the two "2" and the two "4"?
 
  • #3
>4000000 and even number .. The 1st digit must always 4 and 5 and the last digit must be 2 or 4 only to make it even numbers. I tried using the same method i did in question (a) but i can't get the answer .
 
  • #4
Mohd Taqi said:
>4000000 and even number .. The 1st digit must always 4 and 5 and the last digit must be 2 or 4 only to make it even numbers.
Right.

How did you solve (a)?
 
  • #5
I solved a like this . (3.6.5.4.3.2.1)/2!2!2!=270
All possible outcomes = (7!)/2!2!2!=630
P(A)=270/630=3/7

But when i try to solve b , i didnt get the answer
 
  • #6
Mohd Taqi said:
I solved a like this . (3.6.5.4.3.2.1)/2!2!2!=270
All possible outcomes = (7!)/2!2!2!=630
P(A)=270/630=3/7

But when i try to solve b , i didnt get the answer
mfb said "repetitions not allowed" means no two consecutive digits are the same. (Seems an odd way of expressing that constraint, but I have no other interpretation to offer.). You do not seem to have taken that into account.

Edit: or maybe it means further repetitions are not allowed, so e.g. not three 1s. In which case your answer is correct.
 
  • #7
ImageUploadedByPhysics Forums1424592581.622452.jpg
its like this . I don't understnd
 
  • #8
haruspex said:
mfb said "repetitions not allowed" means no two consecutive digits are the same. (Seems an odd way of expressing that constraint, but I have no other interpretation to offer.). You do not seem to have taken that into account.

Edit: or maybe it means further repetitions are not allowed, so e.g. not three 1s. In which case your answer is correct.
I think it makes more sense in the way it is used in post 3, but then it is kind of trivial (otherwise it would not make sense to list some numbers twice). @Mohd Taqi: Can you please clarify what "no repetition" means?

The last image is too small to be readable.
 
  • #9
Without repitition means our choices get reduced right .? Eg : five numbers 12345 . If can pick 3 nmbrs only, then we can't pick 333 or 444 something like tht
 
  • #10
Okay, so "1" has to appear exactly twice and so on. Good.
 
  • #11
Do you have another way to solve this ?
 
  • #12
The approach you showed in post 5 is good, and can be extended to the point where you have conditions on both the first and the last digit. It is useful to consider the cases 4xxxxxx and 5xxxxxx separately.

And the picture is still small to see what you did.
 
  • #13
mfb said:
It is useful to consider the cases 4xxxxxx and 5xxxxxx separately.
... and maybe break those further into two cases according to the final digit.
mfb said:
And the picture is still small to see what you did.
yes, it's unreadable on my screen, and appears to be missing something at the start.
 

Related to How Many Unique 7-Digit Numbers Can Be Formed Using Specific Repeated Digits?

What is the difference between permutation and combination in probability?

Permutation and combination are both methods used in probability to determine the number of possible outcomes. However, permutation takes into account the order of the outcomes, while combination does not. In permutation, the order of the outcomes matters, while in combination, only the combinations of outcomes matter.

How do you calculate the number of possible permutations?

The number of possible permutations can be calculated by using the formula nPr = n!/(n-r)!, where n is the total number of items and r is the number of items being selected. This formula takes into account the order of the outcomes.

What is the probability of a specific outcome in a permutation?

The probability of a specific outcome in a permutation is determined by dividing the number of ways that outcome can occur by the total number of possible outcomes. For example, if there are 5 possible outcomes and we are interested in one specific outcome, the probability would be 1/5 or 20%.

How do you use permutation to solve real-world problems?

Permutation is often used in real-world problems to determine the number of possible arrangements or combinations of a given set of items. For example, it can be used to determine the number of possible seating arrangements at a dinner party or the number of possible combinations of lottery numbers.

What is the relationship between permutation and other concepts in probability?

Permutation is closely related to other concepts in probability such as combinations, combinations with repetition, and binomial coefficients. These concepts all involve determining the number of possible outcomes based on different conditions and constraints.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
6
Views
815
  • Precalculus Mathematics Homework Help
Replies
20
Views
1K
  • Precalculus Mathematics Homework Help
Replies
18
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
879
  • Precalculus Mathematics Homework Help
Replies
8
Views
527
  • Precalculus Mathematics Homework Help
Replies
1
Views
2K
  • Precalculus Mathematics Homework Help
Replies
34
Views
3K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
  • Precalculus Mathematics Homework Help
Replies
19
Views
2K
Replies
4
Views
2K
Back
Top