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## #1 2009-04-10 19:46:59

soha
Real Member
Registered: 2006-07-07
Posts: 2,530

### it is solved ..still..

This is a solved question in my book. But i did'nt understand it. I have written the solution also here .. so pls help me in this especially in the underlined part ...
Question:-

How many 3-digit numbers can be formed by using the digits 3,4,5 and 6 without repetitions? How many of these are even?

Solution:- 1)
There are 4 digits 3,4,5 and 6
Number of 3 digit numbers =

=4× 3 ×2
= 24

2nd part of the question

boxes:-                        H         T         U
2ways  3 ways  2 ways
2)

Consider 3 -blank boxes
Step 1) First fill up the units place.This place can be filled by any one of the digits 4 or 6 in 2 ways.

Step2) After filling this place, three digits are left out. So the tens place can be filled in 3 ways.

Step3)Similiarly, the hundreds place can be filled in the remaining 2 digits in 2 ways.
By the fundamental principle,all the 3 places can be filled in 2× 3× 2 =12 ways

"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!"
- David O. McKay

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## #2 2009-04-10 19:50:39

soha
Real Member
Registered: 2006-07-07
Posts: 2,530

### Re: it is solved ..still..

soha wrote:

Solution:- 1)
There are 4 digits 3,4,5 and 6
Number of 3 digit numbers =   its not ^4P3 that 3 is below P                                      =4× 3 ×2
= 24

2nd part of the question

boxes:-                        H         T         U
2ways  3 ways  2 ways
2)

how did they get 2 ways ,3 ways, 2 ways?

"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!"
- David O. McKay

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## #3 2009-04-10 20:26:19

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: it is solved ..still..

2 even digits on the right they wanted which are 4 or 6 digits.
Then that leaves 3 ways because of 3 digits to pick from, either
3,4,5 or 3,5,6 if 4 was used.  Then after that, there are only
2 digits left over to pick from for the leftmost digit because
there was just 3 left over before picking the middle one.
That's the gist of it anyway.

igloo myrtilles fourmis

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## #4 2009-04-10 20:45:45

soha
Real Member
Registered: 2006-07-07
Posts: 2,530

### Re: it is solved ..still..

John E. Franklin wrote:

2 even digits on the right they wanted which are 4 or 6 digits.
Then that leaves 3 ways because of 3 digits to pick from, either
3,4,5 or 3,5,6 if 4 was used.  Then after that, there are only
2 digits left over to pick from for the leftmost digit because
there was just 3 left over before picking the middle one.
That's the gist of it anyway.

wat is this right ...left????

i really did'nt understand . pls explain again..

"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!"
- David O. McKay

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## #5 2009-04-10 23:03:05

soha
Real Member
Registered: 2006-07-07
Posts: 2,530

### Re: it is solved ..still..

thanks a lot.... i got some help from my mom and because of u , i understood it..

thank u ....

"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!"
- David O. McKay

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## #6 2009-04-11 03:04:00

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: it is solved ..still..

Oh, that's terrific.  You're welcome.

igloo myrtilles fourmis

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## #7 2009-04-23 18:29:42

jaisyjose
Guest

### Re: it is solved ..still..

can u find out the puzzle
26-63=1
u can shift just 1 digit to any side all the other digit should be in their respective places and answer should be correct

## #8 2009-04-23 18:37:26

smiyc86
Member
Registered: 2009-03-19
Posts: 78

### Re: it is solved ..still..

hi jaisy,

won't the ques be

26 - 63 = -1

I love Maths and Music ... dunno which more

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## #9 2009-04-24 06:47:10

simron
Real Member
Registered: 2006-10-07
Posts: 237

### Re: it is solved ..still..

The way I would have done the original problem is:
We have 4 numbers that we wish to arrange in a number without repetitions. So we have 4x3x2=24 ways to do this. To count the number of even numbers, we must deal with the restrictions first. So we have 2 ways to have the units place, 3 ways to have the tens place, and 2 ways to have the hundreds place. Therefore, we have 3x2x2=12 ways to arrange the number so that it is even.
Does this make sense?

Linux FTW

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## #10 2009-04-25 22:36:46

soha
Real Member
Registered: 2006-07-07
Posts: 2,530

### Re: it is solved ..still..

how come four numbers?

"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!"
- David O. McKay

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## #11 2017-08-17 17:39:36

Monox D. I-Fly
Member
Registered: 2015-12-02
Posts: 857

### Re: it is solved ..still..

Because there are four digits available.

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