Question:
How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3.
[This Question was asked as TITA]
How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3.
[This Question was asked as TITA]
[This Question was asked as TITA]
Updated On: Sep 17, 2024
- 462
- 513
- 315
- 295
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The Correct Option is C
Approach Solution - 1
Consider four blanks
__ __ __ __
7 is in thousand place, then 3 can be placed in any of the 3 places in 3 ways. Remaining two blanks can be filled with remaining eight digits in \(^8P_2\) ways. The number of numbers that have 7 is in thousand place is \(3×^8P_2=168\)
Thousand place cannot be 0,7 and 3, it can be filled with remaining 7 digits in 7 ways. In remaining 3 blanks, 7 and 3 can be arranged in 3 ways. Fourth blank can be filled in 7 ways. The number of numbers that are formed where 7 and 3 is not in thousand place is \(7×3×7=147\).
Hence total required numbers = \(168+147 = 315\)
So, the correct option is (C): \(315\)
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Approach Solution -2
According to the question,
Case 1: When 7 is at the left extreme
In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)
Hence, total ways 3(8)(7)= 168
Case 2: When 7 is not at the extremes
So, there are 3 cases possible and the remaining two places can be filled in 7(7) ways.
(Note: 0 can't come on the extreme left)
Hence, total 3(7)(7)=147 ways
Therefore, Total ways 168+147=315 ways
Case 1: When 7 is at the left extreme
In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)
Hence, total ways 3(8)(7)= 168
Case 2: When 7 is not at the extremes
So, there are 3 cases possible and the remaining two places can be filled in 7(7) ways.
(Note: 0 can't come on the extreme left)
Hence, total 3(7)(7)=147 ways
Therefore, Total ways 168+147=315 ways
Correct Option is (C): 315 ways
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