11 Plus Exams Forum

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 ?

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

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.

You haven't really explained that ....

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.

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.

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.

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.

Definitely not primary!

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!)