three-digit numbers are to be formed from 1,2,3,4 and 5. if repetition is not allowed how many can be formed? what is the probability of selecting an odd number from the numbers formed in the above? pls guys help on this, thanks.
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Originally Posted by lawochekel three-digit numbers are to be formed from 1,2,3,4 and 5. if repetition is not allowed how many can be formed? what is the probability of selecting an odd number from the numbers formed in the above? pls guys help on this, thanks. Let's denote number of three-digit numbers as N . Then : $\displaystyle N=\frac{5!}{(5-3)!}=\frac{5 \cdot 4 \cdot 3 \cdot 2!}{2!}=60$
First digit can be any of 5 numbers, then second any of 4, then third any of 3. Hence answer is 5x4x3=60 Probability number ends in 1 is 1/5. Same for 3 and 5. Hence probability odd is 3/5
60
2 evens and 3 odds...so simply 3/5 will be odd.
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