# Probability with permutation

1. Feb 21, 2015

### Mohd Taqi

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

2. Feb 21, 2015

### Staff: Mentor

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. Feb 21, 2015

### Mohd Taqi

>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 cant get the answer .

4. Feb 21, 2015

### Staff: Mentor

Right.

How did you solve (a)?

5. Feb 22, 2015

### Mohd Taqi

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. Feb 22, 2015

### haruspex

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. Feb 22, 2015

### Mohd Taqi

its like this . I dont understnd

8. Feb 22, 2015

### Staff: Mentor

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. Feb 22, 2015

### Mohd Taqi

Without repitition means our choices get reduced right .? Eg : five numbers 12345 . If can pick 3 nmbrs only, then we cant pick 333 or 444 something like tht

10. Feb 22, 2015

### Staff: Mentor

Okay, so "1" has to appear exactly twice and so on. Good.

11. Feb 22, 2015

### Mohd Taqi

Do you have another way to solve this ?

12. Feb 22, 2015

### Staff: Mentor

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. Feb 22, 2015

### haruspex

... and maybe break those further into two cases according to the final digit.
yes, it's unreadable on my screen, and appears to be missing something at the start.