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How many 4-digit even numbers are there with distinct digits ?
A. 2250
B. 2296
C. 2520
D. 4499
E. 4500
A. 2250
B. 2296
C. 2520
D. 4499
E. 4500
raj2812
If the last digit is 0: xxx0 9*8*7*1
If the last digit is is 2, 4, 6 or 8: xxx(2, 4, 6, 8) 8*8*7*4
9*8*7*1 + 8*8*7*4 = 2296.
Answer: B.
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Posts: 20
two cases are possible.
case 1: if unit digit is 0
the no of 4 digit numbers statifying the condition are : 9*8*7*1 = 504
case2: if unit digit is either 2, 4, 6, 8
the no of 4 digit numbers statifying the condition are : 8*8*7*4 = 1792
total = 2296
case 1: if unit digit is 0
the no of 4 digit numbers statifying the condition are : 9*8*7*1 = 504
case2: if unit digit is either 2, 4, 6, 8
the no of 4 digit numbers statifying the condition are : 8*8*7*4 = 1792
total = 2296
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How many 4-digit even numbers are there with distinct digits ?
Constraint:
Unit digits should be even numbers.
Known information:
4 Digit even numbers end with 0,2,4,6 and 8.
Solution:
1. If the last digit is 0, then number of possible digits is 9*8*7*1 = 504
2. If the last digit is 2,4,6,or 8, then number of possible digits is 8*8*7*4 = 1792
Total number of distinct digits = 504+1792 = 2296
Ans: B
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Constraint:
Unit digits should be even numbers.
Known information:
4 Digit even numbers end with 0,2,4,6 and 8.
Solution:
1. If the last digit is 0, then number of possible digits is 9*8*7*1 = 504
2. If the last digit is 2,4,6,or 8, then number of possible digits is 8*8*7*4 = 1792
Total number of distinct digits = 504+1792 = 2296
Ans: B
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raj2812
4- digit even numbers: Unit Digit is always even - 0,2,4,8
Also note that Thousands Digit should not be ZERO.
If Unit digit is ZERO - 9*8*7 *1 = 504.
9 - because we have 9 digits to choose from excluding zero)
8 - because we have 8 digits to choose after we have fixed UNIT and THOUSANDS digit.
7 - because we have 7 digit to choose after we have fixed UNIT, HUNDREDS, THOUSANDS digit.
If UNIT digit is 2. - 8 * 8* 7* 1 = 448
8 - because we have 8 digits to choose from excluding zero and UNIT digit 2.
8 - because we have 8 digits to choose after we have fixed UNIT and THOUSANDS digit.
7 - because we have 7 digit to choose after we have fixed UNIT, HUNDREDS, THOUSANDS digit.
Similarly for UNIT digit 4,6,8 - for each one of those we will get 448.
Total = 504 + 4*448 = 2296.
Ans - B
Why can't we calculate all 4 digit numbers with distinct digits and divide by 2? Wouldn't the half of the numbers be even?
I did: (9*9*8*7)/2 = 2268. Can anyone please help find the flaw in this approach. I assumed half of the total 4 digit numbers with distinct integers will anyway be even.
I did: (9*9*8*7)/2 = 2268. Can anyone please help find the flaw in this approach. I assumed half of the total 4 digit numbers with distinct integers will anyway be even.
urvashis09
hi urvashis09
the units digit has only 5 options per se. i.e set of even numbers 0,2,4,6,8.
i think you meant (9 * 8 * 7 * 5 = 2520), which is a trap answer, because this product will contain cases where zero has been counted at units place and at the hundereds place/tens place also.
we dont have to count such cases...therefore we take the case of zero at the units place seperately. just try to mull over it, it shall click!!
hope this is of some help.
raj2812
Hi, I think I did not explain my reasoning well in the previous post. So, I calculated ALL the 4 digit numbers with distinct digits by 9 options for first digit (as total options are from 1 to 9), 9 for second digit (as total options are from 0 to 9 i.e., but one is already taken), 8 for third digit (as total options are from 0 to 9 i.e. 10 but two are already taken) and 7 for fourth (as total options are from 0 to 9 i.e. 10 but 3 are already taken). Can you please tell the flaw?
So total number of possible 4 digit numbers with all distinct digits = 9*9*8*7. And half of these numbers will be even, therefore (9*9*8*7)/2 should give the number of even numbers.
urvashis09
Gotcha!! no we cannot do this because this would include the cases where the number "zero" would have been counted at multiple places.
Let me try explaining how..For eg consider any number that follows all the conditions given (i.e 4 digit even number) but ends with zero, lets say "1230".
Now according to your method, if zero has already been taken as the last even digit at the units place, it is again being considered for the second digit (hundreds)/the third digit (tens)...i.e 9 x 9 x 8 x 7. whereas once the zero is used at the unit digits' place it cannot be used at any other place. So the hundreds place will have 8 options (and not 9) and the tens place will have 7 options (and not 8).
Therefore, there are two cases involved in this question and we take both of them seperately, i.e. either 0 can come at the end or 2, 4, 6 and 8 will come at the end. Both the cases are giving different possibilities. Let us explore both the cases:
_ _ _ ( 0 ) + _ _ _ (2, 4, 6 or 8)
Case 1: 9 x 8 x 7 x 1
We have to use zero at the fourth place and there are 9 numbers to choose for three places without any restrictions.
Case 2 : 8 x8 x 7 x 4
Total = 9 x 8 x 7 x 1 + 8 x 8 x 7 x 4 = 2296.
Hope this is of some help!
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raj2812
Since the number has to even, the last digit can only be 0, 2, 4, 6 and 8. However, since the first digit can’t be 0, we will divide this into two cases: 1) the last digit is 0 and 2) the last digit is 2, 4, 6 or 8.
Case 1: The last digit is 0.
The last digit has only 1 choice since it is 0. The first digit has 9 choices, second digit 8 choices and third digit 7 choices. Therefore, there are 9 x 8 x 7 x 1 = 504 such numbers.
Case 2: The last digit is is 2, 4, 6 or 8.
The last digit has 4 choices. The first digit has 8 choices (since it can’t be 0 or the same as the last digit), second digit 8 choices (since it can’t be same as the first or last digit, but it can be 0) and third digit 7 choices. Therefore, there are 8 x 8 x 7 x 4 = 1792 such numbers.
Therefore, there are a total of 504 + 1792 = 2296 such numbers.
Answer: B
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Posts: 13
Bunuel VeritasKarishma mikemcgarry chetan2u or any experts. Kindly help me.
Units digit -> can be filled in 5 ways ( 0,2,4,6,8)
Thousands digit -> can be filled in 8 ways (excluding 0 and the digit selected for units place)
Hundreds digit -> can be filled in 8 ways (excluding digits selected for units and thousands digit)
Tens digit -> can be filled in 7 ways (excluding digits selected for units, hundreds, thousands digits)
Therefore total number of ways = 5*8*8*7= 2240
Please correct where I am going wrong.
Units digit -> can be filled in 5 ways ( 0,2,4,6,8)
Thousands digit -> can be filled in 8 ways (excluding 0 and the digit selected for units place)
Hundreds digit -> can be filled in 8 ways (excluding digits selected for units and thousands digit)
Tens digit -> can be filled in 7 ways (excluding digits selected for units, hundreds, thousands digits)
Therefore total number of ways = 5*8*8*7= 2240
Please correct where I am going wrong.
pallavichsk
The units digit can be selected in 5 ways 0 or 2/4/6/8.
If the units digit is 0, there are 9 options for thousands digit. But if the units digit is 2/4/6/8, there are 8 options for thousands digit.
So how you proceed after selecting the units digit depends on what the units digit is. Hence you need to take two cases.
Units digit is 0 -
Thousands digit in 9 ways, hundreds in 8 ways and tens in 7 ways. Total = 9*8*7
Units digit is 2/4/6/8 - (4 ways)
Thousands digit in 8 ways, hundreds in 8 ways and tens in 7 ways. Total = 4*8*8*7
Total = 2296
Answer (B)
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raj2812
Alternate approach:
(even 4-digit integers with distinct digits) = (all possible 4-digit integers with distinct digits) - (odd 4-digit integers with distinct digits)
All possible options:
Number of options for the thousands digit = 9 (Any digit but 0)
Number of options for the hundreds digit = 9 (Any of the remaining 9 digits)
Number of options for the tens digit = 8 (Any of the remaining 8 digits)
Number of options for the units digit = 7 (Any of the remaining 7 digits)
To combine these options, we multiply:
9*9*8*7
Odd options:
Number of options for the units digit = 5 (Must be odd)
Number of options for the thousands digit = 8 (Any of the remaining 9 digits but 0)
Number of options for the hundreds digit = 8 (Any of the remaining 8 digits)
Number of options for the tens digit = 7 (Any of the remaining 7 digits)
To combine these options, we multiply:
5*8*8*7
Thus:
even options = (9*9*8*7) - (5*8*8*7) = (81*56) - (40*56) = (integer with units digit of 6) - (integer with a units digit of 0) = integer with a units digit of 6
Only B is viable.
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Posts: 13
Thanks a lot Karishma. I can understand this now.
VeritasKarishma
Bunuel
Can you explain why for the case when the last digit is 2, 4, 6, 8 you're multiplying 8*8*7*4 instead of 9*8*7*4?
roshaun25
Because when the last digit is 2, 4, 6, or 8, the first digit cannot be that same digit (digits must be distinct) and it also cannot be 0, so there are only 8 choices for the first digit, not 9. After the first and last digits are fixed, the next digit also has 8 choices: it can be any digit except those two already used, and 0 is allowed here if it has not been used yet.
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