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Exercise 6.1 · Q2

Q.Evaluate n!(n−r)!\dfrac{n!}{(n-r)!}, when n=6,r=2n = 6, r = 2

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✓ Free question

Evaluate the standard permutation-style expression n!(n−r)!\dfrac{n!}{(n-r)!} at n=6n=6, r=2r=2, by cancelling the shared factorial tail.

For integers n≥r≥0n \ge r \ge 0:

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

i.e. the product of the rr largest factors of n!n! — this expression is in fact nPr^{n}P_r, the number of ways to arrange rr items chosen from nn distinct items.

  1. Substitute the given values n=6n=6, r=2r=2: 6!(6−2)!=6!4!\dfrac{6!}{(6-2)!} = \dfrac{6!}{4!}.
  2. Expand 6!=6×5×4!6! = 6\times5\times4! so the trailing 4!4! cancels with the denominator: 6×5×4!4!=6×5\dfrac{6\times5\times4!}{4!} = 6\times5.
  3. Multiply: 6×5=306\times5 = 30.

Self-check: Compute directly — 6!=7206! = 720, 4!=244! = 24, 720/24=30720/24 = 30. ✓ Matches.

✓Final answer

6!4!=30\dfrac{6!}{4!} = 30

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