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

AI Thread Summary
The discussion focuses on calculating the number of unique 7-digit numbers that can be formed from the digits 1, 1, 2, 2, 4, 4, and 5 without repetition. Participants clarify that "no repetition" means no two consecutive digits can be the same, and they explore how to determine the probability of selecting a number greater than 4,000,000 and even. The first digit must be 4 or 5, while the last digit must be 2 or 4 to ensure the number is even. Various methods and calculations are proposed, with some confusion over the interpretation of "no repetition" and how it affects the total outcomes. The conversation emphasizes the need for clear definitions and systematic approaches to solve the probability questions effectively.
Mohd Taqi
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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
 
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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"?
 
>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 .
 
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)?
 
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
 
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.
 
ImageUploadedByPhysics Forums1424592581.622452.jpg
its like this . I don't understnd
 
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.
 
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.
 
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