How many 3 digit numbers can be formed from the digits 12345 if the digits Cannot be repeated?
So basically, I attempted this question as- There are 4 numbers and 3 places to put in the numbers: In the ones place, any 4 numbers can be put, so there are 4 choices in the ones place. Similarly for the tens and the hundreds place. So, the total choices are, by multiplication principle- $$4*4*4=64$$ And well and good, this was the answer. But what if I reversed the method? So I take some particular numbers, like $1,2,3$ and say that, well, $1$ can go in $3$ places, $2$ in $2$ places and $3$ in $1$ place, so by multiplication principle, there are $6$ ways of forming a $3$-digit number with $1,2,3$. But there are $4$ different numbers. So the number of $3$-number combinations are- $(1,2,3)$,$(1,2,4)$,$(1,3,4)$,$(2,3,4)$. Each can be arranged in $6$ ways, so we get $24$ ways totally. So why is my answer different here? Solution : (i) When repetition of digits is allowed:
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How many 3Hence, the total number of required numbers= 9×9×8=648.
How many 5 digit numbers can be formed from 12345 if repetition is not allowed?Not a single five-digit prime number can be formed using the digits 1, 2, 3, 4, 5(without repetition). This is because if one adds the digits, the result obtained will be = 1 + 2 + 3 + 4 + 5 = 15 which is divisible by 3.
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