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]
- 462
- 513
- 315
- 295
The Correct Option is C
Approach Solution - 1
We are to form 4-digit numbers using non-repeating digits from 0 to 9 such that each number contains digits 7 and 3.
Let us analyze two cases:
Case 1: 7 is in the thousand's place
Then the number looks like: 7 _ _ _
Now, we need to place 3 in one of the remaining 3 positions. This can be done in 3 ways.
After placing 7 and 3, we are left with 8 digits (excluding 7 and 3). We now need to fill the other two blanks using these 8 digits, which can be done in:
\( ^8P_2 = 8 \times 7 = 56 \) ways
So, total numbers in this case:
\( 3 \times 56 = 168 \)
Case 2: 7 is not in the thousand's place
Thousand's place cannot be 0, 7, or 3, so we can place any of the remaining 7 digits there.
This can be done in 7 ways.
Now, among the 3 remaining places, we place 7 and 3 in 3 positions taken 2 at a time — that is,
\( ^3P_2 = 6 \) but since we are not distinguishing order yet, and only need to select 2 positions to place 7 and 3 (with no repetition), it's:
\( \text{Number of ways to place 7 and 3} = 3 \times 1 = 3 \)
Now, the remaining 1 blank (after fixing 7 and 3) can be filled from the remaining 7 digits (excluding the 3 used digits), so:
\( 7 \) ways
So total numbers in this case:
\( 7 \times 3 \times 7 = 147 \)
Total required numbers:
\( 168 + 147 = \boxed{315} \)
Correct option: (C) 315
Approach Solution -2
Case 1: When 7 is at the thousand's place (leftmost digit)
In that case, 3 can occupy any of the remaining three places.
The remaining two places can be filled with any two digits from the set {0,1,2,4,5,6,8,9}, which has 8 digits.
So, the total number of such numbers is:
\(3 \times {}^8P_2 = 3 \times 8 \times 7 = 168\)
Case 2: When 7 is not at the thousand's place
Then the thousand's place can be filled with any digit except 0, 3, and 7 ⇒ total of 7 choices.
Out of the remaining 3 positions, we must place both 7 and 3 in any 2 positions (in \(3\) ways), and the last digit can be filled with any of the remaining 7 digits.
So, total number of such numbers is:
\(7 \times 3 \times 7 = 147\)
Total required numbers = \(168 + 147 = 315\)
Correct Option is (C): \(\boxed{315}\)
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