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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  [#permalink]

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New post 21 Sep 2018, 09:36
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Question Stats:

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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!
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Re: How many 5 digit numbers can be formed where the digit  [#permalink]

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New post 21 Sep 2018, 09:56
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Probus wrote:
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!


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

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Brent
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Re: How many 5 digit numbers can be formed where the digit  [#permalink]

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New post 21 Sep 2018, 10:08
Probus wrote:
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!



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
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Re: How many 5 digit numbers can be formed where the digit  [#permalink]

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New post 21 Sep 2018, 10:08
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

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Re: How many 5 digit numbers can be formed where the digit  [#permalink]

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New post 21 Sep 2018, 10:36
chetan2u wrote:
Probus wrote:
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!



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


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

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Re: How many 5 digit numbers can be formed where the digit  [#permalink]

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New post 21 Sep 2018, 18:43
Probus

Couldn't understand the half and greater concept. Cam across such problems before but still didn't get the concept. Please explain.
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Re: How many 5 digit numbers can be formed where the digit  [#permalink]

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New post 21 Sep 2018, 21:30
DaniyalAlwani wrote:
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
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Re: How many 5 digit numbers can be formed where the digit &nbs [#permalink] 21 Sep 2018, 21:30
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