How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
The digits that are given are 0, 1, 3, 5, 7 and 9. Any number that is divisible by 10 must contain 0 at the end of that number. Thus fix 0 at the units place of the 6 digit number. Then 5 digits are left which will be rearranged in such a way that no digit is repeated. It can be done when all the vacant spaces are occupied by the remaining 5 digits taken all at a time.
The formula for the permutation is,
P n r = n ! ( n − r ) !
Substitute 5 for n and 5 for r in the above formula.
P 5 5 = 5 ! ( 5 − 5 ) ! = 5 ! 0 ! = 5 ! 1 = 5 !
The formula to calculate n ! is defined as,
n ! = 1 × 2 × 3 × ⋯ × ( n − 1 ) × n
Thus the factorial of 5 is,
5 ! = 5 × 4 × 3 × 2 × 1 = 120
Thus, the number of ways the 6 digit numbers are formed is 120.
The digits that are given are 0, 1, 3, 5, 7 and 9. Any number that is divisible by 10 must contain 0 at the end of that number. Thus fix 0 at the units place of the 6 digit number. Then 5 digits are left which will be rearranged in such a way that no digit is repeated. It can be done when all the vacant spaces are occupied by the remaining 5 digits taken all at a time.
The formula for the permutation is,
Substitute 5 for n and 5 for r in the above formula.
The formula to calculate
Thus the factorial of 5 is,
Thus, the number of ways the 6 digit numbers are formed is 120.
