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Question: How many 5 digit telephone numbers can be constructed using the digits 0 to 9 if each number starts ...

How many 5 digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Explanation

Solution

We use the method of combinations to fill out the 3 remaining digit places of the 5 digit number where the first two places are fixed as 67. Exclude the 2 choices 6 and 7 and calculate the number of ways to fill 3 places using the combination method.

  • Combination is given by nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} , where n is the total number of available objects and r is the number of objects we have to choose.
  • Factorial terms open up as n!=n(n−1)!n! = n(n - 1)!

Complete step-by-step answer:
We are given the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9
So, there are 10 digits in total
We have to form a 5 digit number where first two positions are fixed as 6 and 7
So, we exclude the digits 6 and 7 and we now have available digits 0, 1, 2, 3, 4, 5, 8 and 9
So, now we have 8 digits to choose from
We calculate the number of ways to fill the third, fourth and fifth position and we deduct one choice which has been made after each position.
We use the method of combination nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
Whenever the value of r is 1 we can write nC1=n!(n−1)!1!^n{C_1} = \dfrac{{n!}}{{(n - 1)!1!}}
From the formula of factorial we know n!=n(n−1)!n! = n(n - 1)!
So we can write nC1=n(n−1)!(n−1)!^n{C_1} = \dfrac{{n(n - 1)!}}{{(n - 1)!}}
On cancelling same terms from numerator and denominator we get
nC1=n^n{C_1} = n … (1)
Third Position:
We have total number of letters to choose from as 8, so value of n=8n = 8
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill first position is 8C1^8{C_1}
Using the equation (1) we get 8C1=8^8{C_1} = 8
So, the number of ways to fill the first position is 8.
Fourth Position:
We have total number of letters to choose from as 7, so value of n=7n = 7
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill second position is 7C1^7{C_1}
Using the equation (1) we get 7C1=7^7{C_1} = 7
So, the number of ways to fill the second position is 7.
Fifth Position:
We have total number of letters to choose from as 6, so value of n=6n = 6
We have to choose one letter from the given letters, so value of r=1r = 1
Therefore, number of ways to fill third position is 6C1^6{C_1}
Using the equation (1) we get 6C1=6^6{C_1} = 6
So, the number of ways to fill the third position is 6.
Number of total numbers formed from the digits 0 to 9 such that number stars from 67 are given by multiplication of number of ways to fill each position.
Therefore, total number of numbers formed =1×1×8×7×6 = 1 \times 1 \times 8 \times 7 \times 6
On multiplying the values we get number of total numbers =336 = 336

∴\therefore Total number of numbers that can be formed is 336.

Note:
Many students make the mistake of writing the number of digits to choose from in third, fourth and fifth position as the same i.e. as 8 which is wrong as we are given there is no repetition which means the digit once allotted will not be allotted to another position again.