Question

The number of 5-digit numbers (no digit is repeated) that can be formed by using the digits 0, 1, 2,…..,7 is
A.1340
B.1860
C.2520
D.5880
E.3200

Answer 1

Hint: Here, we have to find the number of 5-digit numbers. We will use the concept of Permutation. We will find the number of ways of choosing each digit of the 5- digit number using the formula of permutation. Then we will add all the ways to find the required answer. Permutation is a method of arranging all the members of a set into some sequence or order.
Formula Used:
 Permutation is given by the formula \[{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}\]

Complete step-by-step answer:
We have to find the number of 5-digit number
First, we will choose one number from the given 8 digits but zero should be excluded as the first digit. Since the number should be a 5-digit number if zero is assigned as a first digit, then the number becomes a 4-digit number.
So, the number of possibilities in the first digit \[n = 7\] and \[r = 1\] .
Number of ways choosing a first digit of a number \[ = {}^7{P_1}\]
Now using the formula \[{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}\], we get
\[ \Rightarrow \] Number of ways choosing a first digit of a number \[ = \dfrac{{7!}}{{(7 - 1)!}} = \dfrac{{7!}}{{6!}}\]
\[ \Rightarrow \] Number of ways choosing a first digit of a number\[ = 7\]
Second, we have to choose one number from the 7 digits since repetition is not allowed and zero can be included as the second digit.
So, the number of possibilities in the second digit \[n = 7\] and \[r = 1\] .
Number of ways choosing a second digit of a number\[{}^7{P_1} = \dfrac{{7!}}{{(7 - 1)!}} = \dfrac{{7!}}{{6!}}\]
\[ \Rightarrow \] Number of ways choosing a second digit of a number\[{}^7{P_1} = 7\]
Next, we have to choose one number from the 6 digits since repetition is not allowed.
So, the number of possibilities in the third digit \[n = 6\] and \[r = 1\].
Number of ways choosing a third digit of a number\[{}^6{P_1} = \dfrac{{6!}}{{(6 - 1)!}} = \dfrac{{6!}}{{5!}}\]
\[ \Rightarrow \] Number of ways choosing a third digit of a number\[{}^6{P_1} = 6\]
Next, we have to choose one number from the 5 digits since repetition is not allowed.
So, the number of possibilities in the fourth digit \[n = 5\] and \[r = 1\].
Number of ways choosing a fourth digit of a number \[{}^5{P_1} = \dfrac{{5!}}{{(5 - 1)!}} = \dfrac{{5!}}{{4!}}\]
\[ \Rightarrow \] Number of ways choosing a fourth digit of a number \[{}^5{P_1} = 5\]
Next, we have to choose one number from the 4 digits since repetition is not allowed.
So, the number of possibilities in the fifth digit \[n = 4\] and \[r = 1\].
Number of ways choosing a fifth digit of a number \[{}^4{P_1} = \dfrac{{4!}}{{(4 - 1)!}} = \dfrac{{4!}}{{3!}}\]
\[ \Rightarrow \] Number of ways choosing a fifth digit of a number \[{}^4{P_1} = 4\]
So, the total number of 5-digit number \[ = 7 \times 7 \times 6 \times 5 \times 4 = 5880\]
Therefore, the number of 5-digit numbers that can be formed by using the digits 0, 1, 2,…..,7 is 5880.
Hence, option D is the correct answer.

Note: We might make a mistake by using the combinations instead of permutation. Usually we use permutations in arranging the numbers, alphabets etc., whereas combination is the selection of team, etc. Permutation is used for the list of data where the order of the data matters and the combination is used for a group of data where the order of data doesn’t matter. Here we had to arrange the digits without repetition so we used permutation.