Go to navigation
It is currently Wed Nov 14, 2018 5:31 am

All times are UTC




Post new topic Reply to topic  [ 14 posts ]  Go to page 1, 2  Next
Author Message
PostPosted: Thu Jun 28, 2018 6:05 pm 
Offline

Joined: Thu Jun 28, 2018 5:37 pm
Posts: 1
How many 5 digit no s with a value less than 95000 Can be made using digits 7,2,0,5,9 ?
How to do these type of problems easily ?


Top
 Profile  
 
PostPosted: Thu Jun 28, 2018 6:28 pm 
Offline

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 16014
How many 5 digit no s with a value less than 95000 Can be made using digits 7,2,0,5,9?

You need to work logically -

Start with 9 second digit can't be 5 or 7 so 92057, 92075, 92507, 92570, 92705, 92750

then start with 7 and repeat

The alternative is to think about all the numbers and subtract the ones that are too big.

xxxxx can fill the first in 5 ways, the second in 4 etc


Top
 Profile  
 
PostPosted: Thu Jun 28, 2018 9:37 pm 
Offline

Joined: Sun Aug 13, 2017 12:50 pm
Posts: 8
The total numbers can be made from 5 different digits is 5*4*3*2*1=120.
95000 is near the top value end.
So just cut off the numbers bigger than 95000.
These numbers include numbers start with 95???, which include 3*2*1=6 numbers.
Also include numbers start with 97???, which also include 3*2*1=6 numbers.
So together, there are 12 numbers bigger than 95000.
Then, there are 120-12=108 numbers smaller than 95000.

However in this question, it means only 5-digits numbers. So we have to cut off the 4-digits numbers as well, which is numbers starts with 0, there are 4*3*2*1=24 numbers.

So ,108-24=84.


Last edited by Mitumum on Thu Jun 28, 2018 9:52 pm, edited 1 time in total.

Top
 Profile  
 
PostPosted: Thu Jun 28, 2018 9:41 pm 
Offline

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 16014
You haven't really explained that ....


Top
 Profile  
 
PostPosted: Thu Jun 28, 2018 10:04 pm 
Offline

Joined: Sun Aug 13, 2017 12:50 pm
Posts: 8
By the way, I do not think this should be a 11+ question. It is not tricky, but the concept is a bit hard to explain to kids. Children are more likely to do it the hard way which will take ages.


Top
 Profile  
 
PostPosted: Thu Jun 28, 2018 10:25 pm 
Offline

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 16014
You still haven't explained where the 5 x 4 x 3 x 2 x 1 came from - as you say this is not an 11+ approach.


Top
 Profile  
 
PostPosted: Thu Jun 28, 2018 11:16 pm 
Offline

Joined: Sun Aug 13, 2017 12:50 pm
Posts: 8
The only easy way is to use permutation and combination formula,that's where the 5*4*3*2*1 come from.
But I do not how to explain this concept to kids, maybe use examples? I do not think this is in primary math curriculum.
Once they know the formula, then it is not so tricky. But if they don't, it is really hard.


Top
 Profile  
 
PostPosted: Fri Jun 29, 2018 9:28 am 
Offline

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 16014
As you say this is certainly not in the curriculum and, as a maths specialist, I think it's difficult to teach 'properly' the difference between permutations and combinations.

I'd go with my first approach. Once you've done part of the listing you can guess the rest.


Top
 Profile  
 
PostPosted: Fri Jun 29, 2018 10:59 am 
Offline

Joined: Mon Jun 18, 2007 2:32 pm
Posts: 7275
Location: East Kent
Definitely not primary!


Top
 Profile  
 
PostPosted: Fri Jun 29, 2018 3:43 pm 
Offline

Joined: Fri Mar 10, 2017 4:06 pm
Posts: 506
Mitumum wrote:
The only easy way is to use permutation and combination formula,that's where the 5*4*3*2*1 come from.
But I do not how to explain this concept to kids, maybe use examples? I do not think this is in primary math curriculum.
Once they know the formula, then it is not so tricky. But if they don't, it is really hard.


I would say look at each digit in turn.

There are 5 possibilities for the first digit 0, 2, 5, 7, 9.
Then for each of these there are 4 possibilities for the second digit (because you've already used up one of the 5 for the first digit).
Then similarly for each of these there are 3 possibilities for the 3rd digit, 2 for the 4th digit and finally only one possibility for the last digit as there's only one digit left by then as you've already used up the other 4 by then.

That's where the 5 * 4 * 3 * 2 * 1 comes from.

And then as you say you need to exclude all the cases beginning 97***, 95*** and 0****, which can be worked out in the same way.

So basically its 5! - 4! - 2(3!)


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 14 posts ]  Go to page 1, 2  Next

All times are UTC


Who is online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
CALL 020 8204 5060
   
Privacy Policy | Refund Policy | Disclaimer | Copyright © 2004 – 2018