How many 5 digit even numbers can be made if: repetition is not allowed and 4 must be included. Use the indirect method to do this: So basically i believe , you have to find the # of ways, 5 digit even #s that can be made, then subtract it from even numbers that do not include 4. but i don't know how to act on my plan, how can i find how many 5 digit even numbers
there are. and how can i find the number of ways that do not include 4 , but are even numbers?
grapz said: How many 5 digit even numbers can be made if: repetition is not allowed and 4 must be included. The first question is what is considered a five digit number? Is 4=00004 considered such a number? If so the answer is \[\displaystyle \left[ 5 \right]\left[ {10^4 } \right]\]. If the first digit must be at least 1,
example 10400, then the answer is \[\displaystyle \left[ {10^4 } \right] + \left[ 8 \right]\left[ 4 \right]\left[ {10^3 } \right]\].
hm the defination of a 5 digit number in this case would be for the first number to be at least one. but, repetition is not allowed. i believe the method u shown here allows repetition.
grapz said: a 5 digit number in this case would be for the first number to be at least one. but, repetition is not allowed. In that case the answer is \[\displaystyle \left[ 9 \right]\left[ 8 \right]\left[ 7 \right]\left[ 6 \right] + \left[ 8 \right]\left[ 4 \right]\left[ 8 \right]\left[ 7 \right]\left[ 6 \right]\]. Count
those beginning with a 4 and add those that do not.
thanks. So this is now what i did. The question was, how many 5 digit even numbers can be made if repetition is not allowed and the number four must be included. so i did 8 x 8 x 7 x 6 x 4 - 7 x 7 x 6 x 5 x 3 = 6342. but the answer is 7686. 8x8x7x6x4- thats the number of 5 digit even numbers, with no repetition. and 7x7x6x5x3 is when 4 is excluded.pka
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grapz said: The question was, how many 5 digit even numbers can be made if repetition is not allowed and the number four must be included. Are you saying that the answer is 7686? There are 3024 five-digit numbers that begin with
the digit 4 with no repetition.
the answer is 7686.
If so, then your answer key is just wrong!
The correct answer is 13776.
So can you understand how many more there are?
Re: There are how many 4-digit even numbers if the first digit cannot be a [#permalink]
Bunuel wrote:
There are how many 4-digit even numbers if the first digit cannot be a zero?
A. 3,600
B.
3,645
C. 4,500
D. 4,999
E. 5,000
Hi,
FIRST, lets write down the restrictions
1] these are four digit even numbers, so first digit can be 1 to 9, and the units digit can be 2,4,6,8, or 0..
2] any number of repetition can be done..
Now, lets work out a solution on these restrictions
1] units digit= any of five even digits.. 0,2,4,6,8..
2] at tens and hundreds position any of the ten digits can be placed ..
3] at thousands
place, any of the 10 except 0 can be placed...
the ways= 5*10*10*9= 4500
another way
1] what are the different ways without any restriction =10*10*10*10=10,000..
2] out of these ways in which first digit is 0...= 1*10*10*10=1000..
3] Remaining= 10000-1000=9000
4] Half of 9000 will be even and other half odd.. so even = 9000/2=4500..
ans C
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