How many 5 digit numbers can be formed where the digit

I applied perms and Combs with all possible solutions and got the answer as C. But the approach by GMATPrepNow seems less time consuming. Good one

Posted from my mobile device

Hi,
Although I got the ansy but I couldn't understand the concept that half shud be greater and half shud be smaller!

Can u pls explain?

Thanks and regards

Posted from my mobile device

Probus

Couldn't understand the half and greater concept. Cam across such problems before but still didn't get the concept. Please explain.

HeyDaniyalAlwani

I am not expert but i can try.

Well multiple approached for this question chetan2u has shown .
Here is another ( lenghty)
so lets see in which cases we get solution

case 1:
Say we have 2 in units place then we can only have 1 in tens place . and the remaining places can be filled in 3! ways . So it would be 3!*1 ways
case 2:
Say we have 3 in units place then we can only have 1 or 2 in tens place . and the remaining places can be filled in 3! ways . So it would be 3!*2 ways
case 3:
Say we have 4 in units place then we can only have 1 or 2 or 3 in tens place . and the remaining places can be filled in 3! ways . So it would be 3!*3 ways
case 4:
Say we have 5 in units place then we can only have 1 or 2 or 3 or 4 in tens place . and the remaining places can be filled in 3! ways . So it would be 3!*4 ways
Now combining all case we have
3!*1+3!*2+3!*3+3!*4
so we have
3!( 1+2+3+4)
3!(10)
6(10)
60.



Regarding your question

Just take a very small sample . say we are forming digits using 1,2 3, 4 . What could be total possible ways 4!.
But did you recognize that Say starting with 1 there would be 6 numbers, . So each digit contributes 6 different numbers.

1234
1243
1324
1342
1423
1432

Did you notice a pattern here . so of 6 numbers that start with 1 half have Tens's digit greater than unit's digit and other half have unit's digit greater than tens's digit.
Now in a set of 6 half have last two digits where units place is greater than tens place.

So in a set of 24, we would have 12 where we have unit's digit greater than ten's digit.

In our case we have 120 ways where each digit contributes 24 different numbers. So ineach set of 24 numbers 12 will have unit's digit greater than ten's digit.

12* 5= 60

In case you want to visualize.


2134
2143
2314
2341
2413
2431

3412
3421
3214
3241
3124
3142

4312
4321
4213
4231
4123
4132


Does this help
Probus

1, 2, 3, 4 5 (a b c d e:: e > d)

possible1: 1 & 2,3 4,5
Possible2: 2 & 3,4,5
Possibel3: 3 & 4,5
Possible4:4 & 5


3*2*1*1*4 = 24
1*3*3*2*1 = 18
1*2*3*2*1 = 12
1*1*6 = 6
total = 60

The _ represent digits: _ _ _ X(X+N)
We have 5*4*3 options for the first 3 digits
The last digits can then only be filled in 1 way (2 numbers remain, one must be larger than the other).
Therefore
5*4*3*1=60

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.