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

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 =

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

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

= 242nd part of the question

2)

boxes:- H T U

2ways 3 ways 2 ways

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

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

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

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

thank u ....

- David O. McKay

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

Oh, that's terrific. You're welcome.

**igloo** **myrtilles** **fourmis**

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**jaisyjose****Guest**

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

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

hi jaisy,

won't the ques be

26 - 63 = -1

I love Maths and Music ... dunno which more

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

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

how come four numbers?

- David O. McKay

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**Monox D. I-Fly****Member**- Registered: 2015-12-02
- Posts: 886

Because there are four digits available.

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