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sara
how many four digit numbers begin and end with an even numbers? Can we solve this problem through arithmetic progression? how?
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jiju
Ans: $2000$

Solution 1: Using Arithmetic Progression

Consider the numbers $2000,2002,\cdots,2998$
Count of these numbers
$=\dfrac{2998-2000}{2}+1=500$

Similarly,  each of $(4000,\cdots,4998),$ $(6000,\cdots,6998),$ $(8000,\cdots,8998)$ gives $500$ numbers.

So, required count
$500+500+500+500=2000$

Solution 2 (using combinatorics)

First digit can be any of $(2,4,6,8)$
Second digit can be any of  the $10$ digits
Third digit can be any of  the $10$ digits
Fourth digit can be any of $(0,2,4,6,8)$

Required count
$=4×10×10×5=2000$
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