How many 4-digit numbers, each greater than 1000

Question

How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

A. 147
B. 189
C. 315
D. 420
E. 441

(adapted from gmatfree)

VipinChaurasia · Answer

bhai koi solve karo yar ise

av1901 · Answer

Let us start by taking the cases where 7 can come before 3

A) 3 in units place

_ _ _ 3

Now in this case, 7 can appear in any of the remaining digits

A1) If 7 appears in tens place:

_ _ 7 3

We have 7 options for thousands place (distinct required and that place cannot have 0 since we are dealing with four digit numbers) and 7 options for the hundreds place = 7*7 = 49

A2) If 7 appears in hundreds place:

_ 7 _ 3

Same as above = 49

A3) If 7 appears in thousands place:

7 _ _ 3

We have 8 options for hundreds place and 7 options for tens place = 56 (in this case 0 cannot come in thousands place since this case has the digit 7 there)

B) 3 in the tens place

_ _ 3 _

Now, in this, we can have the digit 7 either in the hundreds place or in the thousands place

B1) 7 in the hundreds place

_ 7 3 _

We have 7 options for the thousands place and 7 remaining options for the units place = 7*7 = 49

B2) 7 in the thousands place

7 _ 3 _

We have 8 options for the hundreds place and 7 options for the units place = 8*7 = 56

C) 3 in the hundreds place

_ 3 _ _

Now, here we can only have 7 in the thousands place

C1) 7 in the thousands place

7 3 _ _

We have 8 options for the tens place and 7 options for the units place = 8*7 = 56

So we have considered all cases and the sum of all cases is = 49*3 + 56*3 = 315

Answer - C

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Kinshook · Answer

Asked: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

AB73: 7*7*1*1 = 49 cases
A7B3: 7*1*7*1 = 49 cases
7AB3: 1*8*7*1 = 56 cases
73AB: 1*1*8*7 = 56 cases
7A3B: 1*8*1*7 = 56 cases
A73B: 7*1*1*7 = 49 cases

Total 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3 = 49*3 + 56*3 =147 + 168 = 315

IMO C

chetan2u · Answer

Let the number be ABCD

(I) METHOD 1
If 0 is also there, then 0 can take any 3 places - B, C or D. 
7 can take any of the remaining 3 places and 3 can take any of the remaining 2 places. The final place can be taken by any of the remaining 7 digits - (1,2,4,5,6,8,9)
Total: 3*3*2*7=126, half of which will have 7 and 3 in a particular order.

If 0 is not there, then 7 can take any 4 places and 3 can take any of the remaining 3 places, while the remaining 2 places can be taken by remaining 7 digits in 7*6 ways 
Total: 4*3*7*6=504, half of which will have 7 and 3 in a particular order.

Answer = 63+252 = 315

(II) METHOD 2
Let A be 7, that is 7BCD. 
Fix 3 in any of the remaining 3 places- B, C or D. The remaining two places can be fixed by remaining 8 digits in 8*7 ways. 
Thus, 3*8*7=168
Now, let B be 7, that is A7CD. 
Fix 3 in any of the remaining 2 places- C or D. The remaining two places can be fixed by remaining 8 digits in 7*7 ways, as A can take any 7 digits (other than 0) and the other empty space can take any 7 digits (including 0). 
Thus, 2*7*7=98
Next, let C be 7, that is AB7D. 
You can fix 3 at D only. The remaining two places can be fixed by remaining 8 digits in 7*7 ways, as A can take any 7 digits (other than 0). 
Thus, 1*7*7=49

Total = 168+98+49 = 315

How many 4-digit numbers, each greater than 1000

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