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Worked Examples · Example 19

Q.How many 4-digit numbers can be formed using the digits 1 to 9 if no digit is repeated.

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✓ Free question

Choosing and arranging 4 distinct digits (order matters, forms a number) out of 9 available digits is a direct permutation.

[!FORMULA]

nPr=n!(n−r)!^{n}P_r = \dfrac{n!}{(n-r)!}: number of ways to arrange rr objects chosen from nn distinct objects.

  1. Digits available: 1,2,…,91,2,\dots,9, so n=9n=9; digits needed for a 4-digit number: r=4r=4.
  2. Since no digit is repeated and every digit 11–99 is non-zero, there is no leading-digit restriction to worry about separately.
  3. Number of 4-digit numbers =9P4=9!(9−4)!=9!5!=9×8×7×6= {}^{9}P_4 = \dfrac{9!}{(9-4)!} = \dfrac{9!}{5!} = 9\times8\times7\times6.
  4. Compute: 9×8=729\times8 = 72; 72×7=50472\times7 = 504; 504×6=3024504\times6 = 3024.
  5. Self-check: 9!/5!=362880/120=30249!/5! = 362880/120 = 3024 ✓.
✓Final answer

30243024 four-digit numbers

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