find out how many groups of 3 numbers can be formed from the given numbers
Answers
Well 3 digit numbers formed by 1,2,3 can be repetitive or non repetitive type
If repetition of any number is allowed:
Then 3³ = 3x3x3 = 27 possibilities
Suppose xyz is 3 digit number then
For x we have 3 possibilities, for y 3 possibilities and for x again 3 possibilities if repetition is allowed so possible numbers = 3x3x3 = 27
Numbers are
111, 112, 113, 121, 122, 123, 131, 132, 133
211, 212, 213, 221, 222, 223, 231, 232, 233
311, 312, 313, 321, 322, 323, 331, 332, 333
Total = 27
If repetition of any number is not allowed:
If repetition is not allowed then
Take example of xyz
Now for x we have 3 possibilities
For y we cannot use the number already used for x so (3–1) = 2 possibilities
For z we cannot use the number already used for x and y so we left with (3–2) = 1 possibility
So total 3 digit number possibilities without repetition is 3x2x1 = 6
Numbers are
123, 132, 213, 231, 312, 321
Total = 6
We can also calculation it through permutation :
nPr = n! / (n-r)!
Now we have to form 3 digit number so r=3
We have 3 numbers available so n=3
So
3P3 = 3!/ (3–3)! = 3x2x1 / 0! = 6 /1 = 6
Or if you have required outcome = available resource as in our case both required number = 3 and available digits = 3
We can use directly 3! = 3x2x1 = 6
Hope it will help
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