4-digit Number

[kavya001]

Kindly help in sloving these/

How many even 4-digit numbers can be formed by using digit 0-9 so that no two nos are repeated.

A) 2296
B) 2396
C) 2444
D) 2456
E) 2486

[12rk34]

Total possible numbers = 7*8*9*5 = 2520
This includes the numbers with '0' at thousands place.
Such numbers will be = 7*8*4 = 224
So required nos.= 2520 - 224
= 2296
Hence A.

[kavya001]

Thanks Buddy.

[Meggieski]

I'm not sure how you got the second part? For ones with 0 at the thousands.. shouldn't it be (1*9*8*7)/2 ?

[12rk34]

Originally Posted by Meggieski

I'm not sure how you got the second part? For ones with 0 at the thousands.. shouldn't it be (1*9*8*7)/2 ?

Since the resulting numbers are even, unit digit will be any of the four choices i.e. 2,4,6 or 8.
Tens digit will be one of the 8 remaining numbers (since zero is already at thousands place ) and hundreds digit will be one of the remaining 7 numbers.

[arindambanerji]

Can you help me understand where am I going wrong with this approach

For the thousand place we can have any of the 9 numbers excluding 0
For the hundred place we can have the rest of the 9 numbers including 0
For the tens place we can have the rest of the 8 numbers
& For the units place we can have the rest of the 7 numbers

if there are equal number of odd and even numbers, then
(9*9*8*7)/ 2 = 2268

Where did I go wrong?

[12rk34]

Originally Posted by arindambanerji

Can you help me understand where am I going wrong with this approach

For the thousand place we can have any of the 9 numbers excluding 0
For the hundred place we can have the rest of the 9 numbers including 0
For the tens place we can have the rest of the 8 numbers
& For the units place we can have the rest of the 7 numbers

if there are equal number of odd and even numbers, then
(9*9*8*7)/ 2 = 2268

Where did I go wrong?

I don't think we can presume that odd and even numbers will be equally distrbuted. Please remember that it is a discrete rather than a continuous group of numbers. For example 1120, 1224 etc. will not be included in the list. Now let us see why even numbers will not be equal to odd ones.
For four digit numbers will be like 1xxx, 2xxx, 3xxx, ...9xxx
Now if a digit occupies thousands place it can not be repeated at any other place. For example 2xxx will not have digit '2' at any of the remaining three places, marked as 'x'. So 2xxx will not have any even number like 2xx2.
This is true for 3xxx where we will not have any odd number like 3xx3.
Similar is the case for 4xxx, 5xxx, ... etc.
Now 1xxx, 3xxx, 5xxx, 7xxx and 9xxx will not have odd numbers with 1,3,5,7 or 9 respectively as unit digit.
Similarly 2xxx, 4xxx, 6xxx and 8xxx will not have even numbers with 2,4,6 or 8 respectively as unit digit.
Obviously there are 5 such cases for odd numbers and 4 such cases for even numbers. Therefore the odd and even numbers will be different.

[rajuksj]

Good explanation 12rk34

[kavya001]

Originally Posted by kavya001

Kindly help in sloving these/

How many even 4-digit numbers can be formed by using digit 0-9 so that no two nos are repeated.

A) 2296
B) 2396
C) 2444
D) 2456
E) 2486

Some more question on the same concept
1. How many even 4-digit numbers can be formed by using digit 0-9 which are divisible by 4 & no two nos are repeated.
a) 336 b)784 c)1120 d) 1804 e)1936

How many even 4-digit numbers can be formed by using digit 0-9 so that it contains exactly 3 distinct digits?
a)1944 b)3240 c)3850 d) 3888 e)4216

[12rk34]

Q1:
There will be total 24 numbers divisible by 4 which will have last two digits = 4*1(=04) to 24*24
Out of these 2 nos. can be ignored (44 & 88).
So the last two digits of the numbers will range from 4*1 to 4*24 = 22 numbers. Now 4 digit numbers without any repetition will be of two types:
(a) where one of the last 2 digts will contain zero.(04,08,20,40,60,80)
Total such numbers will be = 8*7*6 = 336
(b) where last 2 digits are non-zero.
Total such numbers = 7*7*16 = 784
Total such numbers will be = 336+784
= 1120
Hence C.

Q2: Such numbers where one of digits will be repeated are in 3 categories:
(a) where only zero is repeated:Zero can take only three positions in 3 ways.
Such numbers will be = 3*9*8 = 216
(b) Non-zero digits (1-9) are repeated that can take any 2 of last 3 positions:
Such numbers will be = 9*8*8*3 = 1728
(b) Non-zero digits (1-9) are repeated that can take first and any one of last 3 positions:
Such numbers will be = 9*9*8*3 = 1944
Total numbers will be = 216+1728+1944
= 3888
Hence D.