calculate how many different numbers can be formed by taking one, two or three digits from the digits 8, 5, 2. How many of these will be odd numbers and greater than than 250? no repetition allowed. i have worked out the first part and i found 15, by using permutation. now i need help on the second part of the question.
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i have worked out the first part using permutation and i found 15. how many of these 15 numbers will be odd umbers and greater than 250 is my main concern – Gibbons Muselitata Sep 12 '16 at 12:39
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It make a difference whether repeated digits are OK, – coffeemath Sep 12 '16 at 12:41
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Greater than $250$ - all $3$ digit numbers. Odd - all numbers ending with $5$. So there are two such numbers: $285$ and $825$. – barak manos Sep 12 '16 at 12:44
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@Crostul: "no repetition allowed". – barak manos Sep 12 '16 at 12:44
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Not following...just listing all the cases appears to work fine. Clearly you need all three digits to get over $250$, so there are only $6$ cases to look at. – lulu Sep 12 '16 at 12:46
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A number is odd if the digits at unit's place / one's place is odd right ?
So starting with one digit number , possible odd cases is with number 5
With two digit numbers , possible odd cases are 2 , this is because the unit's place needs to have a 5 so now you are left with two choices for tens place , so total odd numbers are 2 * 1 = 2
With three digit numbers there are again two possibilities after fixing 5 at units place , tens and hundreds place can have two cases , so total odd numbers in this case are 2
So In total odd numbers are 5 , Now from here you can solve the remaining part how many out of these will be greater than
