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How many numbers can be formed from 1, 2, 3, 4, 5, ( without repetition), when the digit at the unit's place must be greater than that in the ten's place?

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closed as off-topic by jvdhooft, Théophile, M. Winter, Leucippus, Shailesh Jun 19 '18 at 0:31

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  • $\begingroup$ Please have a look at how to ask a good question. $\endgroup$ – Théophile Jun 18 '18 at 22:25
  • $\begingroup$ You can explicitly count this yourself on paper. If there are too many, you can try with $1,2,3,4$ or $1,2,3$ and see if you notice a pattern... $\endgroup$ – Jair Taylor Jun 18 '18 at 22:35
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Hint: how many numbers can be formed at all, and what proportion of those have the digit in the units place bigger than that in the tens place?

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total methods without any restriction $= 5\cdot4\cdot3\cdot2\cdot1 = 120$

Now for units place to be greater than tens place , only one method out of the set of two will be correct.

Ex$:$ $12$ & $21$ only one will be valid so answer will be $\dfrac{120}{2} = 60$

So, $60$ numbers can be formed using all of $1,2,3,4,5($without repetition$),$when the digit at the units place must be greater than that in the tenth place.

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we take 1 as in the tenth place digit then in units place 2,3,4 and 5 can only be used. so the first 3 digits from left can be parmuted using any 3 digits. So number of possible ways are 3! total ways=4*3! for 2,3*3! for 3,2*3! for4,1*3! therefore total numbers will be=(4+3+2+1)*3!=60

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