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Exercise 6.2 · Q5

Q.Evaluate n!(n−r)!\dfrac{n!}{(n-r)!}, when

(i) n=6n = 6, r=2r = 2
(ii) n=9n = 9, r=5r = 5.
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The expression n!(n−r)!\frac{n!}{(n-r)!} counts the number of ways to arrange rr objects from nn distinct objects. For (i) n=6,r=2n=6, r=2: 30; for (ii) n=9,r=5n=9, r=5: 15120.

Understanding Factorial Division

When we divide one factorial by another, we're not meant to compute each factorial separately and then divide—that would be inefficient and miss the elegant cancellation built into the structure. The expression n!(n−r)!\frac{n!}{(n-r)!} represents the product of rr consecutive integers starting from nn and counting downward.

Why? Because n!=n×(n−1)×(n−2)×⋯×2×1n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1, and (n−r)!=(n−r)×(n−r−1)×⋯×2×1(n-r)! = (n-r) \times (n-r-1) \times \cdots \times 2 \times 1. When we divide, everything from (n−r)(n-r) down to 11 cancels out, leaving us with:

n!(n−r)!=n×(n−1)×(n−2)×⋯×(n−r+1)\frac{n!}{(n-r)!} = n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)

This is precisely rr factors, starting at nn and stepping down.

Tip

Count the factors: you need exactly rr terms starting from nn. For 6!4!\frac{6!}{4!}, you get 6×5=26 \times 5 = 2 factors (since 6−4=26-4=2).


(i) When n=6n = 6 and r=2r = 2

We need to evaluate 6!(6−2)!=6!4!\frac{6!}{(6-2)!} = \frac{6!}{4!}.

  1. Identify the cancellation point: We're dividing 6!6! by 4!4!, so everything from 4!4! downward cancels.

  2. Write out the remaining factors:

6!4!=6×5×4!4!=6×5\frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} = 6 \times 5

  1. Compute the product:

6×5=306 \times 5 = 30


(ii) When n=9n = 9 and r=5r = 5

We need to evaluate 9!(9−5)!=9!4!\frac{9!}{(9-5)!} = \frac{9!}{4!}. …

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