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Miscellaneous Examples · Example 53

Q.How many numbers greater than 2000000 can be formed using the digits 1, 3, 0, 3, 2, 3, 2?

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✓ Free question

Using all 7 given digits {0,1,2,2,3,3,3}\{0,1,2,2,3,3,3\} once each to form 7-digit numbers, a number exceeds 2,000,0002{,}000{,}000 exactly when its leading digit is 2 or 3 (since a 7-digit number starting with 1 is under 2 million, and one starting with 0 is not a valid 7-digit number).

Number of distinct arrangements of nn items with repetitions of sizes p,q,r,…p,q,r,\dots: n!p! q! r!⋯\dfrac{n!}{p!\,q!\,r!\cdots}. Here the 7 digits are {0,1,2,2,3,3,3}\{0,1,2,2,3,3,3\}: digit 22 repeats twice, digit 33 repeats three times.

  1. Given digits: 0,1,2,2,3,3,30,1,2,2,3,3,3 — 7 digits total, with 22 appearing twice and 33 appearing three times.
  2. A 7-digit number formed from all 7 digits exceeds 2,000,0002{,}000{,}000 if and only if its first digit is 22 or 33 (a leading 00 is invalid as a 7-digit number, and a leading 11 gives a number <2,000,000<2{,}000{,}000; no digit exceeds 33 here so nothing needs separate handling beyond this split).

Case A: number starts with 2

3. Remaining 6 digits to arrange: {0,1,2,3,3,3}\{0,1,2,3,3,3\} (one 22 used, one 22 and three 33's remain).

4. Arrangements =6!3!=7206=120=\dfrac{6!}{3!}=\dfrac{720}{6}=120 (only the digit 33 repeats, 3 times, among these 6).

Case B: number starts with 3

5. Remaining 6 digits to arrange: {0,1,2,2,3,3}\{0,1,2,2,3,3\} (one 33 used, two 22's and two 33's remain).

6. Arrangements =6!2! 2!=7204=180=\dfrac{6!}{2!\,2!}=\dfrac{720}{4}=180.

  1. Total numbers greater than 2,000,0002{,}000{,}000 =120+180=300=120+180=300.
✓Final answer

Numbers greater than 2,000,0002{,}000{,}000 =120+180=300= 120+180 = 300

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