DaniyalAlwani wrote:
ProbusCouldn't understand the half and greater concept. Cam across such problems before but still didn't get the concept. Please explain.
Hey
DaniyalAlwaniI 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.
123412431324134214231432Did 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.
213421432314234124132431341234213214324131243142431243214213423141234132Does this help
Probus
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