smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]

smurfy:

The numbers are integers, so only 2 and 2 qualify as answers.

Calculusf(x):
hmmm ,for numbers greater than 4000(it can be 4,5 or 6 digits)if 4digits,we can see 4 and 5,and for the number to end with even,any of 0,2 or 4 must end the number,also note that the number must not be repeated,then we do it like this,make hole,4 for four digits,5 for five digits etc
4 DIGITS STARTING WITH 5
it can end with 0,2 or 4,then the fourth hole will have 3,it starts with 5,the first hole with have 1,don't forget there are six numbers in all,the last and the first are known therefore it gives 1*4*3*3=36
4 DIGITS STARTING WITH 4
since repeatition is not allowed,then it can end with 0 or 2,then the holes will have 1*4*3*2=24
total=36+24=60
5 DIGITS
it can end with 0,2 or 4.but note that 0 cannot be in the first hole so as not to make the number 4digits again(then only 4 digits can occupy the first hole)...then the holes are 4*4*3*2*3=288...
6 DIGITS
Same as 5 digits...it can end with 0,2 or 4,then 3 can occupy the last hole,and can start with 4 in the first hole coz 0 cannot start the digit...then 4*4*3*2*1*3=288...
TOTAL=288+288+60=636ways

factorial1: 0 and 0 too

Dane17: . (6 digit ending with 0 can be arranged in(cbai)) 5×4×3×2=120 ;(6 digit ending with 2 or 4 cbai) 4×4×3×2=96 each. total six digit =96+96+120=312. total five digit numba is also 312. (4 digit ending with 0 and 2 cbai) 2×4×3= 24 each ; (4 digit ending with 4 cbai) 1×4×3=12 ; total 4 digit= 24+24+12=60. Total numba greater dan 4000 dat can be formed is 60+312+312=684

smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]

smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]

factorial1: Play with the questions below:

2. The height of a cylinder is 10cm and its radius is 4cm. Find the approximate increase in volume when the radius increases to 4.02cm

4. The side of a square is 5cm. How much will the area of the square increase when the side expands by 0.01cm? And lastly

smurfy:

Beautifully correct! Wow! I love your answer. Now, for the sake of others, let me explain the steps below...

smurfy:

Master Calculus has already got 60 for all possible 4-digit numbers, so I'll skip that.

Here goes...

5-DIGIT NUMBERS
Now we need to proceed with caution because of that 0 digit. There are three scenarios to consider:
(a) ends with 0
(b) ends with 2
(c) ends with 4

(A) Ends with Zero:
Possible combination is 5 * 4 * 3 * 2 * 1 = 120

(B) Ends with Two
Possible combination is 4 * 4 * 3 * 2 * 1 = 96

(C) Ends with Four
Possible combination is 4 * 4 * 3 * 2 * 1 = 96

Add these three scenarios to get 312 possible 5-digit numbers.

6-DIGIT NUMBERS
Same three scenarios.

The three combinations (following the same order as above) are:

5*4*3*2*1*1 = 120 (ends with zero),
4*4*3*2*1*1 = 96 (ends with two) and
4*4*3*2*1*1 = 96 (ends with four). This also gives 312.

So the number of ways of forming even numbers greater than 4000 from digits 0 through 5 is 60 + 312 + 312 = 684.

Laplacian: @smurfy
2 ansa ur ques i'll first mak a few observations

1.) d numbers 0,1,2,3,4,5 wen used 2 form numbers greater dan 4000 will contain equal number of even numbers as odd numbers because the last digits can each be in 3 ways; 0,2,4(even) and 1,3,5(odd)

2.)if u are to arrange a set of n numbers and u do not desire 0 to
start d digit, then first arrange d numbers i.e nPr
Then keep zero constant as d first number and arrang d rest i.e
(n-1)P(r-1)
If now u take their differenc
i.e nPr-(n-1)P(r-1)
Then the result is d number of ways of arrangin d numbers without zero appearin as first digit

Now Ur Solution

ARRANGE IN SIX DIGITS
number of ways of arrangin 0,1,2,3,4,5 in six digit witout zero appearin first is
=6P6-5P5=600ways

ARRANGE IN FIVE DIGITS

6P5-5P4=600ways

ARRANGE IN FOUR DIGITS

in dis case 0,1,2,3 MUST not occur as first digit, so number of ways of arrangement
=6P4-4*(5P3)=360-4*60
=120ways

in summary, number of ways of formin digit wit 0,1,2,3,4,5 so dat d number formed is greater than 4000=600+600+120=1,320ways
finally if d numbers formed must be EVEN, then number of ways
=1320/2=660ways

Dane17: . d 6 and 5 digit will not have d same numba of odd and even numba. let's use ur method for 0,1,2. from ur form numba of 3 digit will be 3P3-2P2=4 and numba of even numba would be 4/2=2. But dere are 3 even(210,102,120) and one odd (201 numba.

Laplacian:
i do not claim dat my solution is right but u hav decisively justified my approach...using my approach u correctly predicted d four numbers, but d odd and even numbers are not equal because my FIRST OBSERVATION is not satisfied; 0,2(even) and 1(odd)...
Try it d numbers 0,1,2,3 and c

benbuks: Find the anti-derivative of (cosx +1)^-2

Dane17: . I'm not saying I'm right either I just stated an observation. for 0,1,2,3 u get 10 even (102,120,130,132,210,230,302,310,312,320) and 8 odd (103,123,201,203,213,231,301,321). ur oda formula should be correct as 4P3-3P2=18

smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]

smurfy:

Master Calculus has already got 60 for all possible 4-digit numbers, so I'll skip that.

Here goes...

5-DIGIT NUMBERS
Now we need to proceed with caution because of that 0 digit. There are three scenarios to consider:
(a) ends with 0
(b) ends with 2
(c) ends with 4

(A) Ends with Zero:
Possible combination is 5 * 4 * 3 * 2 * 1 = 120

(B) Ends with Two
Possible combination is 4 * 4 * 3 * 2 * 1 = 96

(C) Ends with Four
Possible combination is 4 * 4 * 3 * 2 * 1 = 96

Add these three scenarios to get 312 possible 5-digit numbers.

6-DIGIT NUMBERS
Same three scenarios.

The three combinations (following the same order as above) are:

5*4*3*2*1*1 = 120 (ends with zero),
4*4*3*2*1*1 = 96 (ends with two) and
4*4*3*2*1*1 = 96 (ends with four). This also gives 312.

So the number of ways of forming even numbers greater than 4000 from digits 0 through 5 is 60 + 312 + 312 = 684.

Laplacian:
thank u very much bro....

Why are d even numbers more dan d odd numbers?....**:ojust thinking:/**

busuyem:


sorry 4 asking this question: why did u repeat 4 twice while dealing with numbers that end with 2 and 4 digits?

busuyem:


sorry 4 asking this question: why did u repeat 4 twice while dealing with numbers that end with 2 and 4 digits?

Laplacian:
pls i'll need u 2 solve it 4 d case in where we are required 2 find all odd numbers greater than 4000 using 0,1,2,3,4,5

godofphysics: Pls, any reasonable workings to dis questn?
Find x&y if x+y=5 &x^x+y^y=31 (i tink d answers r 2&3, i'd b glad if i can get d workings)

rhydex 247:
try nd check the previous pages u will c d working there.

smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]