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Combinatorics - Factorials-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The factorial of a non-negative integer nn, denoted by n!n!, is the product of all positive integers from 11 to nn.

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Special case: By convention, the factorial of zero is defined as 0!=10! = 1.

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The recursive property of factorials is given by n!=n×(n−1)!n! = n \times (n-1)!, which allows for the simplification of complex fractions involving factorials.

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Trailing Zeros: The number of trailing zeros in n!n! is determined by the number of times 55 is a factor in the product n!n!, calculated as ⌊n5⌋+⌊n25⌋+…\lfloor \frac{n}{5} \rfloor + \lfloor \frac{n}{25} \rfloor + \dots.

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Divisibility: If n≥5n \geq 5, then n!n! is always divisible by 1010 and ends in the digit 00.

📐Formulae

n!=n×(n−1)×(n−2)×⋯×3×2×1n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1

0!=10! = 1

n!=n×(n−1)!n! = n \times (n-1)!

n!r!=n×(n−1)×⋯×(r+1)\frac{n!}{r!} = n \times (n-1) \times \dots \times (r+1) (where n>rn > r)

Number of trailing zeros in n!=∑k=1∞⌊n5k⌋\text{Number of trailing zeros in } n! = \sum_{k=1}^{\infty} \lfloor \frac{n}{5^k} \rfloor

💡Examples

Problem 1:

Evaluate the expression 10!7!×3!\frac{10!}{7! \times 3!}.

Solution:

10!7!×3!=10×9×8×7!7!×(3×2×1)\frac{10!}{7! \times 3!} = \frac{10 \times 9 \times 8 \times 7!}{7! \times (3 \times 2 \times 1)} 10×9×83×2×1\frac{10 \times 9 \times 8}{3 \times 2 \times 1} 7206=120\frac{720}{6} = 120

Explanation:

We expanded 10!10! until we reached 7!7! so that we could cancel the 7!7! in the denominator. Then, we expanded 3!3! and simplified the remaining fraction.

Problem 2:

Find the value of nn if (n+1)!(n−1)!=30\frac{(n+1)!}{(n-1)!} = 30.

Solution:

(n+1)×n×(n−1)!(n−1)!=30\frac{(n+1) \times n \times (n-1)!}{(n-1)!} = 30 (n+1)×n=30(n+1) \times n = 30 n2+n−30=0n^2 + n - 30 = 0 (n+6)(n−5)=0(n+6)(n-5) = 0 n=5 or n=−6n = 5 \text{ or } n = -6 Since factorials are defined for non-negative integers, n=5n = 5.

Explanation:

We use the recursive property (n+1)!=(n+1)×n×(n−1)!(n+1)! = (n+1) \times n \times (n-1)! to simplify the ratio, resulting in a quadratic equation.

Problem 3:

Determine the number of trailing zeros in 100!100!.

Solution:

Number of zeros=⌊1005⌋+⌊10025⌋+⌊100125⌋\text{Number of zeros} = \lfloor \frac{100}{5} \rfloor + \lfloor \frac{100}{25} \rfloor + \lfloor \frac{100}{125} \rfloor Number of zeros=20+4+0=24\text{Number of zeros} = 20 + 4 + 0 = 24

Explanation:

To find trailing zeros, we count the number of factors of 55 in the product 100!100!. This is done by dividing 100100 by powers of 55 and summing the integer quotients.