Question

How many 3 -digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) Repetition of the digits is allowed? (ii) Repetition of the digits is not allowed?


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Answer 1

Answer:

(i) 125, (ii) 60

Step-by-step explanation:

We have 5 numbers, and we want to calculate the total number of 3-digit numbers to be formed.

(i) If repetition is allowed.

$[ \ \ \ ] [ \ \ \ ] [ \ \ \ ]$

First digit can be filled with any of the 5 numbers, so no. of ways is 5.

Similarly, for the second digit and third digit, the no. of ways is 5 and 5 respectively (as repetition is allowed).

$[ \ \ 5\ \ ][ \ \ 5\ \ ][ \ \ 5\ \ ]$

So, total number of ways is $5 \times 5 \times 5 = 125$

Or if repetition is allowed, you can do it like $(No.\ of \ digits \ given)^(What \ digit \ no. \ to \ be \ formed)$

Here, its $5^3=125$

(ii) If repetition is not allowed.

$[ \ \ \ ] [ \ \ \ ] [ \ \ \ ]$

First digit can be filled with any of the 5 numbers, so no. of ways is 5.

Now, we have one less number in our stack that can be filled.

Second digit can be filled with any of the 4 numbers, so no. of ways is 4.

Similarly, third digit can be filled with any of 3 numbers, so no. of ways is 3.

So, total number of ways is $5\times4\times3=60$

Or the number of permutations is $^5P_3=60$