Question

plz help ASAP plzzzzzzzzzzzzzz​

Answer 1

Let the 3 digit number be ABC,

where C units place ,B tens place and A hundred place.

(I) When repetition is allowed-

The number of digits possible at C is 5.

as , repetition is allowed then,B & A also 5.

Therefore the total number of possible 3 digit number = 5×5×5=125

(II) When repetition is not allowed-

The number of possible outcomes at C= 5 and then at B=4 and A=3

Therefore, The total number of possible 3-digit numbers=5 × 4 × 3=60

Hope it helps you ❣️☑️☑️

Answer 2

Answer:

Given number of digits from which the number will be formed = n = 4 .(1, 2,3,4)

Number of digits contained in one number =

r = 3 (as the number is three digit number ).

Now,

Number of ways using permutation

(i) when repetition is allowed

when repetition is allowed number of ways

$= {n}^{r} \\ = {4}^{3} \\ = 64 ways$

(ii) when repetition is not allowed

$= \frac{n !}{(n - r)! } \\ = \frac{4!}{(4 - 3)!} \\ = \frac{4!}{1!} \\ = 4! \: \: \: \: \: \: \: (as \: 1! = 1) \\ = 4 \times 3 \times 2 \times 1 \\ = 24ways$