Question

If we have the numbers $1,2,3{\text{ and 4}}$ how many combinations of 4-digit numbers can we make?

Answer 1

We use the concepts of permutations and combinations to solve this problem. Permutations and combinations are all related to selecting and sorting or arranging things, in which we have to consider every single case to get a perfect value. We will also learn how to evaluate these permutations and combinations.

Complete step-by-step solution:
Firstly, consider that there are $m$ different things or objects. And now, to select $n$ things or objects from these, we use combinations, which gives us the number of ways of selecting $n$ objects from $m$ objects. It is represented as ${}^m{C_n}$ and its value is given by ${}^m{C_n} = \dfrac{{m!}}{{n!\left( {m - n} \right)!}}$
And the number of arrangements of $m$ objects taken $n$ at a time is given by permutations and is represented as ${}^m{P_n}$ and its value is given as ${}^m{P_n} = \dfrac{{m!}}{{\left( {m - n} \right)!}}$
For example, if we have four numbers say $1,2,3,4$ , then number of ways of selecting two numbers out of these four is given by ${}^4{C_2}$ and its value is ${}^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} = \dfrac{{4!}}{{2!2!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 6$
And the combinations are $(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)$
Take another example, in which there are five numbers say $1,2,3,4,5$ and we need to arrange these five numbers when taken two at a time. And to do so, we use permutations and it is given by ${}^5{P_2}$ and its value is ${}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} = 20$
And the permutations are $(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(4,1),(4,2),(4,3),(3,1),(3,2),(2,1)$
Now let us solve the given question.
Now, as per question we need to find number 4-digit numbers which can be formed by numbers 1,2,3 and 4
So, to solve this, first we need to select numbers from given numbers and then we need to arrange them.
So, number of ways of selecting four numbers (as we need a 4-digit number) from 1,2,3 and 4 is ${}^4{C_4} = \dfrac{{4!}}{{4!\left( {4 - 4} \right)!}} = \dfrac{{4!}}{{4!0!}} = 1$
And now we need to arrange the numbers of this one combination.
And number of ways of arranging four numbers from 1,2,3 and 4 is ${}^4{P_4} = \dfrac{{4!}}{{\left( {4 - 4} \right)!}} = 4! = 24$
So, we get 24 four-digit numbers formed by numbers 1,2,3 and 4.

Note: Here we get a positive integer as a result of permutations or combinations. If you get a negative value or a fractional value, then your solution has gone wrong in some way. Suppose that we got $a$ combinations, then we need to arrange each and every combination. So, the total number of arrangements will be $a \times ({\text{number of arrangements in one combination)}}$