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 Post subject: Easy method to solve this problemPosted: Thu Jun 28, 2018 6:05 pm

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 ?

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 Post subject: Re: Easy method to solve this problemPosted: Thu Jun 28, 2018 6:28 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 16126
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

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

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 Post subject: Re: Easy method to solve this problemPosted: Thu Jun 28, 2018 9:37 pm

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.
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.

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 Post subject: Re: Easy method to solve this problemPosted: Thu Jun 28, 2018 9:41 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 16126
You haven't really explained that ....

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 Post subject: Re: Easy method to solve this problemPosted: Thu Jun 28, 2018 10:04 pm

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.

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 Post subject: Re: Easy method to solve this problemPosted: Thu Jun 28, 2018 10:25 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 16126
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.

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 Post subject: Re: Easy method to solve this problemPosted: Thu Jun 28, 2018 11:16 pm

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.

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 Post subject: Re: Easy method to solve this problemPosted: Fri Jun 29, 2018 9:28 am

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 16126
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.

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 Post subject: Re: Easy method to solve this problemPosted: Fri Jun 29, 2018 10:59 am

Joined: Mon Jun 18, 2007 2:32 pm
Posts: 7343
Location: East Kent
Definitely not primary!

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 Post subject: Re: Easy method to solve this problemPosted: Fri Jun 29, 2018 3:43 pm

Joined: Fri Mar 10, 2017 4:06 pm
Posts: 546
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!)

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