how many $3$ digit numbers can be formed by $1,2,3,4$, when the repetition of digits is allowed?
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:
No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
[ii] When repetition of digits is not allowed:
No. of ways of choosing first digit = 5
No. of ways of choosing second digit = 4
No. of ways of
choosing thrid digit = 3
Total possible numbers `= 5 xx 4 xx 3 = 60`.
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How many 3-digit numbers can b...
Updated On: 27-06-2022
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Text Solution
Solution : [i] `---`
`= 5*5*5= 125`
[ii] `---`
`5*4*3 = 60`
answer
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