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Permutations and Combinations - Derivation of the formula for nPr

Grade 11CBSE

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

🔑Concepts

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A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.

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The number of permutations of nn distinct objects taken rr at a time, without repetition, is denoted by nPr^nP_r or P(n,r)P(n, r).

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To derive the formula, we consider filling rr vacant places with nn available objects. The first place can be filled in nn ways, the second in (n−1)(n-1) ways, the third in (n−2)(n-2) ways, and so on.

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The rr-th place can be filled in (n−(r−1))(n - (r - 1)) ways, which simplifies to (n−r+1)(n - r + 1) ways.

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By the Fundamental Principle of Counting, the total number of ways is the product: n(n−1)(n−2)...(n−r+1)n(n-1)(n-2)...(n-r+1).

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To express this in factorial notation, we multiply and divide the expression by (n−r)!(n-r)!, leading to the standard formula: nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}.

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The constraints for the formula are 0≤r≤n0 \le r \le n and nn must be a positive integer.

📐Formulae

nPr=n(n−1)(n−2)…(n−r+1)^nP_r = n(n-1)(n-2)\dots(n-r+1) outdoor

nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}

nPn=n!^nP_n = n!

0!=10! = 1

nP0=1^nP_0 = 1

💡Examples

Problem 1:

Evaluate the value of 8P3^8P_3.

Solution:

Using the formula nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}, we substitute n=8n=8 and r=3r=3: 8P3=8!(8−3)!=8!5!^8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} 8P3=8×7×6×5!5!^8P_3 = \frac{8 \times 7 \times 6 \times 5!}{5!} 8P3=8×7×6=336^8P_3 = 8 \times 7 \times 6 = 336

Explanation:

The formula for permutations is applied by calculating the factorial of nn divided by the factorial of (n−r)(n-r) to find the number of ways to arrange 3 objects out of 8.

Problem 2:

Find nn if nP2=42^nP_2 = 42.

Solution:

We know that nP2=n(n−1)^nP_2 = n(n-1). Given n(n−1)=42n(n-1) = 42, we solve the quadratic equation: n2−n−42=0n^2 - n - 42 = 0 (n−7)(n+6)=0(n - 7)(n + 6) = 0 Since nn must be a positive integer, n=7n = 7.

Explanation:

By expanding the permutation formula for r=2r=2, we get a product of two consecutive integers. Solving the resulting quadratic equation gives the value of nn.

Problem 3:

Calculate the difference between 5P5^5P_5 and 5P4^5P_4.

Solution:

First, calculate 5P5^5P_5: 5P5=5!=120^5P_5 = 5! = 120 Next, calculate 5P4^5P_4: 5P4=5!(5−4)!=1201!=120^5P_4 = \frac{5!}{(5-4)!} = \frac{120}{1!} = 120 Difference: 120−1200\begin{array}{r} 120 \\ -120 \\ \hline 0 \end{array}

Explanation:

This example demonstrates that nPn=nPn−1^nP_n = ^nP_{n-1} because dividing n!n! by 1!1! is the same as n!n!.