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Question

Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

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Solution

The 4-digit numbers can be formed in many ways by using the digits 1, 2, 3, 4 and 5. It depends upon the permutation of 5 different digits taken 4 at a time.

The formula to calculate the permutation is,

P n r = n! ( nr )!

Where,n is the number of objects taken r at a time.

Since there is the combination of 5 digits from which 4 are taken at a time. So,substitute 5 for n and 4 for r in the above formula.

P 5 4 = 5! ( 54 )! = 5! 1!

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( n1 )! n!=n( n1 )( n2 )![ n2 ]

The permutation can be written as,

P 5 4 = 5×4×3×2×1! 1! =5×4×3×2 =120

The number of ways of the formation of 4-digit numbers is 120.

An even number is formed when it contains an even digit at its unit’s place. Since 2 and 4 are the even digits, thus the number of possible ways to fill the unit’s place is 2.

Since the repetition of digits is not allowed, so the left three places can be occupied by remaining 4 digits. The 3-digit number can be formed in various ways depending upon the permutation of 4 different digits taken 3 at a time.

The formula to calculate the permutation is,

P n r = n! ( nr )!

Where,n is the number of objects taken r at a time.

Since there is the combination of 4 digits from which 3 are taken at a time. So, substitute 4 for n and 3 for r in the above formula.

P 4 3 = 4! ( 43 )! = 4! 1!

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( n1 )! n!=n( n1 )( n2 )![ n2 ]

The permutation can be written as,

P 4 3 = 4×3×2×1! 1! =4×3×2 =24

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 4-digit even numbers is m×n.

Total even numbers that can be formed is,

2×24=48

Thus, the required number of4-digit even numbers is48.


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