*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,

133211, 212, 213, 221, 222, 223, 231,

232, 233311, 312, 313, 321, 322, 323,

331, 332, 333

*Total = 27

SO, THE CORRECT ANSWER IS LETTER B..,^_^

B. 27

Explanation:

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