X  
View & Edit Profile Sign out
X
Sign in
Google
Facebook
Twitter
Yahoo
LinkedIn
X
Discussion Board
showing 1-1 of 1 (answers : 1, comments : 0),   sorted newest to the oldest
Question
how many four digit numbers begin and end with an even numbers? Can we solve this problem through arithmetic progression? how?
 
(0) (0) Comment Answer
numberspermutations & combinations
2017-01-18 06:50:27 
sara
1 Answer
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
$=4×10×10×5=2000$
 
(0) (0) Comment
2017-01-21 18:21:53 
jiju (Junior Maths Expert, careerbless.com)
Answer this Question

Name
5 + 2 = (please answer the simple math question)
Post Your Answer