How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
The digits are given as 1, 2, 3, 4, 6 and 7.
The total number of digits given is 6. Out of these 6 digits, unit’s place can be occupied only by an even digit because an even number is formed when it contains an even digit at its unit’s place. Since 2, 4 and 6 are the even digits, thus there are only 3 digits required for the unit’s place. The number of possible ways at unit’s place is 3.
Since the repetition of digits is not allowed, so the left two places can be occupied by remaining 5 digits. The 2-digit number can be formed in various ways depending upon the permutation of 5 different digits taken 2 at a time.
The formula to calculate the permutation is,
P n r = n ! ( n − r ) !
Where,n is the number of objects taken r at a time.
Since there is the combination of 5 digits from which 2 are taken at a time. So, substitute 5 for n and 2 for r in the above formula.
P 5 2 = 5 ! ( 5 − 2 ) ! = 5 ! 3 !
To cancel the common factor from numerator and denominator, factorize the bigger term in factorials.
The formula to calculate the factors of a factorial in terms of factorial itself is,
n ! = n ( n − 1 ) ! n ! = n ( n − 1 ) ( n − 2 ) ! [ n ≥ 2 ] ⋮
The permutation can be written as,
P 5 2 = 5 × 4 × 3 ! 3 ! = 5 × 4 = 20
By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the total number of ways of forming 3-digit even numbers is m × n .
The total number of 3-digit number that can be formed is,
3 × 20 = 60
Thus, 60 three-digit even numberscan be formed by the given digits.
The digits are given as 1, 2, 3, 4, 6 and 7.
The total number of digits given is 6. Out of these 6 digits, unit’s place can be occupied only by an even digit because an even number is formed when it contains an even digit at its unit’s place. Since 2, 4 and 6 are the even digits, thus there are only 3 digits required for the unit’s place. The number of possible ways at unit’s place is 3.
Since the repetition of digits is not allowed, so the left two places can be occupied by remaining 5 digits. The 2-digit number can be formed in various ways depending upon the permutation of 5 different digits taken 2 at a time.
The formula to calculate the permutation is,
Where,n is the number of objects taken r at a time.
Since there is the combination of 5 digits from which 2 are taken at a time. So, substitute 5 for n and 2 for r in the above formula.
To cancel the common factor from numerator and denominator, factorize the bigger term in factorials.
The formula to calculate the factors of a factorial in terms of factorial itself is,
The permutation can be written as,
By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the total number of ways of forming 3-digit even numbers is
The total number of 3-digit number that can be formed is,
Thus, 60 three-digit even numberscan be formed by the given digits.
