How many 5 digit numbers can be formed where the digit

Question

How many 5 digit numbers can be formed where the digit at unit's place must be greater than digit in ten's place using digits 1,2,3,4,5 when repetition of digits is not allowed.

(A) 48
(B) 54
(C) 60 
(D) 4! + 3!
(E) 4! * 3!

BrentGMATPrepNow · Accepted Answer

Since repetitions are not allowed, we are essentially arranging the 5 digits (1, 2, 3, 4, and 5) to create the required 5-digit numbers

We can arrange n unique objects in n! ways
 
So, we can arrange the 5 digits in 5! ways (=  120  ways)

Of course not all  120   possible outcomes meet the condition that the units digit must be greater than the tens digit.

However, we should recognize that, if we randomly select one of those  120   possible 5-digit numbers, the likelihood that the  units digit is greater  is THE SAME as the likelihood that the  tens digit is greater .

This means that HALF of the  120   numbers will meet the given condition and HALF of the  120   numbers will NOT meet the given condition.

HALF of  120   = 60

Answer: C

Cheers 
Brent

chetan2u · Answer

two ways..

total ways = 5! 
but we have a restriction in last two digits..  the digit at unit's place must be greater than digit in ten's place  
so HALF will have tens greater and other half units greater
5!/2=3*4*5=60

OR

5 places .. _, _, _, _, _
first one 5 ways, next 4 ways and next 3 ways so 5*4*3*_*_
now last two can be placed in only one way - the bigger of two at ten's and smaller at unit's
so 5*4*3*1*1=60

C

Kritika2008poddar · Answer

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

honneeey · Answer

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

DaniyalAlwani · Answer

Probus

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

Probus · Answer

Hey DaniyalAlwani

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

BSiddartha · Answer

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

SchruteDwight · Answer

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

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How many 5 digit numbers can be formed where the digit

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How many 5 digit numbers can be formed where the digit

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