what is the total number of integers with distinct digits that exceed 5500 and do not contain 0 7 and 9
Answers
The given number is 5500.
To find :
The total number of integers exceeding 5500 with distinct digits other than 0, 7, and 9.
Solution :
• The numerical digits remaining after excluding 0,7, and 9 from from 0 - 9 are :
1, 2, 3, 4, 5, 6, and 8.
• 5500 is a 4-digit integer. Therefore, the integers greater than 5500 will all be either 4-digit numbers or more.
Case 1 : Four-digit integers greater than 5500 :
• The 4-digit integers greater than 5500 will start with either 5, 6, or 8.
(i) For a 4-digit integer starting with 5 :
• If an integer starts with the digit 5, then the digit in the thousands place is fixed as 5.
• The distinct digit in the hundreds place, then, must either be 6, or 8.
Therefore, the possibility of either 6 or 8 to be in the hundreds place is 2.
• The digit in the tens place will be a digit from the remaining 5 digits, and the digit in the units place will be a digit from the remaining 4 digits.
• The total number of possible arrangements for the tens and units place are = 5 × 4
• So, we have the total number of possible integers for a distinct 4-digit number greater than 5500 as 1 × 2 × 5 × 4 = 40
(ii) For a 4-digit integer starting with 6 or 8 :
• The digit in the thousands place can be any of the two numbers, 6 or 8.
• The digit in the hundreds place will be one of the remaining 6 digits - 1, 2, 3, 4, 5, or 8 (for a number starting with 6), and 1, 2, 3, 4, 5, or 6 (for a number starting with 8).
• After a number getting fixed in the hundreds place, the digit in the tens place will be any of the remaining 5 digits, and consequently, the digit in the units place will be any of the remaining 4 digits.
• Therefore, the possible number of integers for a distinct 4-digit number greater than 5500, starting with 6 or 8 are : 2 × 6 × 5 × 4 = 240
Case 2 : Five-digit integers greater than 5500 :
• The integer can start with any of the given 7 digits - 1, 2, 3, 4, 5, 6, or 8.
• The possible number of arrangements for a distinct 5-digit integer with the given 7 digits is : 7 × 6 × 5 × 4 × 3 = 2520
Case 3 : Six-digit integers greater than 5500 :
• The required integer can start with any of the given 7 digits, and there will be total 6 digits in the number.
• Therefore, the possible number of arrangements for a distinct 6-digit integer with the given 7 digits is : 7 × 6 × 5 × 4 × 3 × 2 = 5040
Case 4 : Seven-digit integers greater than 5500 :
• The integer can start with any of the given 7 digits, and there will be total 7 digits in the same.
• Therefore, the possible number of arrangements for a distinct 7-digit number with the given 7 digits is : 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
• It is obvious that integers not more than 7 digits can be formed with the aforementioned 7 digits. Therefore, from the above calculations, the total number of required integers that can be formed are = 40 + 240 + 2520 + 5040 + 5040 = 12,880
Answer - The total number of integers exceeding 5500 with distinct digits other than 0, 7, and 9 is 12,880.