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How many $4$-digit even numbers have all $4$ digits distinct?

1. $2240$
2. $2296$
3. $2620$
4. $4536$

Case 1: Pos1 is for odd digit:

pos1 : 5 option

pos4 : 5 option (even number including 0)

pos2 : 8 option (two numbers already used, no repetition allowed)

pos3 : 7 option

total =

case 2: pos 2 is for even digit

pos1 : 4 option (0 cannot be there as its 4 digit number)

pos4: 4 option (one of 2,4,6,8 already used so 3 left + we have 0 so 4 options)

pos2 : 8 option

pos1 : 7 option

total = 5*5*8*7 + 4*4*8*7 = 1400 + 896 = 2296
I have tried in this way -

No of 4 digit even numbers= Total no of 4 digit even numbers -  Number having 0 at the front

Total no of 4 digit even numbers =last place is filled in 2,4,6,8,0 ie 5 ways. Next place in 9 ways, next in 8 ways and next in 7 ways (go from right)

Number having 0 at the front = last place is filled in 2,4,6,8 ie 4 ways, next place in 8 ways and next in 7 ways and first place is for 0.

Required number of numbers = (7*8*9*5)-(7*8*4)  = 2520-224 = 2296

Same question could be done in one more way.

(All four digit no.) - (All four digit no. that are not even).

= 9*9*8*7 – 8*8*7*5

= 2296

We have to find distinct (non repeating) 4 digit numbers possible.

The question says it’s an even number.

Now consider this

___      ___     ___      _X_

in the place marked X we can put an even number only cause then only 4 digit number will be even.

from 1 to 9 we have even numbers 0, 2, 4, 6, 8

now let’s see cases when X is 2, 4, 6, 8. X with value 0 will be the last case I will discuss.

case 1X = 2

___      ___     ___      _2_

now, in the places that are not filled i.e

_a_      _b_     _c_      _2_

→ a can be filled with 8 digits (1,3,4,5,6,7,8,9)  “0 can’t be filled and 2 can’t be filled” [remember the distinct number property and number is a 4 digit number].

→ b can be filled with 8 digits. Why? it’s simple we can now fill b with any of [0, and remaining 7 digits]

→ c can be filled with 7 digits [obviously].

So total ways = 8*8*7 = 448 ways.

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now, with X = 4,6,8

the method remains same [can check it out]

and therefore total ways = 448*4 = 1792 ways

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the last case when x = 0 is the only one left

so when,           _a_      _b_     _c_      _0_

a can be filled with 9 numbers,

b can be filled with 8 numbers,

c can be filled with 7 numbers

total ways = 9*8*7 = 504 ways

so the grand total ways possible = case when x is 2,4,6,8 + case when x is 0

= 1792 + 504 = 2296 ways

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