# How many numbers can be formed? [closed]

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?

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

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – jvdhooft, Théophile, M. Winter, Leucippus, Shailesh
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• Please have a look at how to ask a good question. – Théophile Jun 18 '18 at 22:25
• 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... – Jair Taylor Jun 18 '18 at 22:35

## 3 Answers

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?

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.

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