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Exercise 6.3 · Q15

Q.How many positive integers greater than 5,000,000 can be formed using the digits 2, 3, 3, 5, 5, 6, 8?

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Using all 7 digits, a number >5,000,000>5{,}000{,}000 needs its leading digit to be 55, 66, or 88; count arrangements of the remaining 6 digits (respecting the repeated 3's and 5's) for each case and add.

Arrangements of nn items with repeated items of counts p1,p2,…p_1,p_2,\ldots =n!p1! p2!⋯=\dfrac{n!}{p_1!\,p_2!\cdots}. Digit multiset: {2,3,3,5,5,6,8}\{2,3,3,5,5,6,8\} — 7 digits, with 33 repeated twice and 55 repeated twice.

  1. Total (unrestricted) 7-digit arrangements of all given digits =7!2! 2!=50404=1260=\dfrac{7!}{2!\,2!}=\dfrac{5040}{4}=1260 (none of the digits is 00, so every arrangement is a valid 7-digit number).
  2. A 7-digit number exceeds 5,000,0005{,}000{,}000 iff its leading digit is ≥5\ge5, i.e. leading digit ∈{5,6,8}\in\{5,6,8\} (digits below 5 available are 2,3,32,3,3).
  3. Leading digit =6=6: remaining 6 digits are {2,3,3,5,5,8}\{2,3,3,5,5,8\} (3 twice, 5 twice); arrangements =6!2! 2!=7204=180=\dfrac{6!}{2!\,2!}=\dfrac{720}{4}=180.
  4. Leading digit =8=8: remaining 6 digits are {2,3,3,5,5,6}\{2,3,3,5,5,6\} (3 twice, 5 twice); arrangements =6!2! 2!=180=\dfrac{6!}{2!\,2!}=180. …
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