Question

In the situation where taking the numbers 123456 how many numbers can you form using 3 digits with no numbers repeated is that a permutation or a combination?

Answer 1

Hint: For the arrangement of objects or numbers, the concept of permutation and combination is used. In the given question, we have to group three of the given six numbers and then we also have to arrange the 3 numbers to form different numbers. So, using the formula to calculate the combination and permutation, we can find out the correct answer to the above question.

Complete step-by-step solution:
We are given numbers 123456, and we have to form numbers using 3 digits with no numbers repeated, selection of 3 numbers out of 6 can be done as –
\[
  ^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} \\
  ^6{C_3} = \dfrac{{6!}}{{3!(6 - 3)!}} = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!(3 \times 2 \times 1)}} \\
  { \Rightarrow ^6}{C_3} = 20 \\
 \]
Now, the arrangement of these three numbers can be done as –
$P = 3! = 3 \times 2 \times 1 = 6$
The number of 3-digit numbers formed is equal to the product of the combination and the permutation –
$20 \times 6 = 120$

Hence 120 numbers can be formed using 3 digits with no numbers repeated.

Note: To arrange something in a group in a specific order, we use permutation but when we have to group something and the order of the elements doesn’t matter then we use the concept of combination. We can also solve the given question theoretically, we have six digits and we have to form a 3-digit number. The first place can be taken by any of the 6 given numbers, as repetition is not allowed so the second place can be taken by 5 numbers and the third place by 4 numbers, so the number of 3-digit numbers formed is $6 \times 5 \times 4 = 120$ .