# How many numbers lying between 10 and 100 can be formed with the help of digits 1,3,5,7,9?

Please help me!! i don't know how resolve this problem. This is exercise from permutations in my school. hope you guys can help me :)

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At worst you could write down a list of all the numbers and count it by hand. If you have a particular question about how to approach this in a non-bruteforce approach, please be more explicit and explain what you tried already. – Asaf Karagila Jan 10 '13 at 12:08

nPr will give you the answer => correct me if i am wrong, I learned probability some 7 years back :)

here n= 5 {1,3,5,7,9} and r= 2

Correction : nPr + n Will be the answer

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${}_nC_r$ counts the number of combinations, which are unordered sets of distinct elements. Thus, you would miss $11$, $33$, etc. – Shaun Ault Jan 10 '13 at 12:14
This does not count $33$ – Amr Jan 10 '13 at 12:14
i edited the answer, it is nPr – Prasanth Bendra Jan 10 '13 at 12:17
answer for it will be 20 – Prasanth Bendra Jan 10 '13 at 12:17
Please correct your answer, or delete it. There's no shame in deleting an answer (I've done so, quite a few times. – amWhy Jan 10 '13 at 14:36

You need to construct all 2-digit number by $\{1,3,5,7,9\}$ so for first digit you have $5$ number, that can select, and for the second too.

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all permutations with repetitions of order 2 from set $\{1,3,5,7,9\}$are $$11,13,31,33,15,51,55,17,71,77,19,91,99$$ $$35,53,37,73,39,93$$ $$57,75,59,95$$ $$79,97$$

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Any number between $10,100$ is a two digit number.

What is the number of ordered pairs that are made up using elements in the set $\{1,3,5,7,9\}$ ?

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