Permutations I think

[Loki123]

I have been given a hint that the answer is not smaller than 1000 and not bigger than 7000. With my way I would get waaaay bigger numbers... Where am I going wrong??

 

[Loki123]

I have been given a hint that the answer is not smaller than 1000 and not bigger than 7000. With my way I would get waaaay bigger numbers... Where am I going wrong?? View attachment 32308

I thought to use combinations but it's still a very large number.

 

[BigBeachBanana]

I thought to use combinations but it's still a very large number.
View attachment 32309

You're assuming that the numbers can't be repeated. Where does it say that?

 

[Loki123]

You're assuming that the numbers can't be repeated. Where does it say that?

If they were to repeat the possibilities would be even bigger.

 

[JeffM]

The question is not well stated. Are the choices 1, 3, 7 and 7, 3, 1 distinct? Is 3, 3, 3 permitted?

If we assume no repetitions are permitted and order is irrelevant, I get 20 odd integers in 1 through 40 and 40 even.

[math]\dbinom{20}{3} * \dbinom{20}{0} + \dbinom{20}{1} * \dbinom{20}{2} =\\ \dfrac{20 * 19 * 18}{3 * 2} + 20 * \dfrac{20 * 19}{2} =\\ 380 + 20 * 190 = 380 + 3800 = 4180.[/math]

 

[Cubist]

No-one has posted the correct result for [imath]\binom{20}{3}[/imath] yet (both post#2 and #5 contain mistakes). I'm sure that @Loki123 can get this right now, with another attempt, since @JeffM probably just mistyped something in the actual calculation.

 

[JeffM]

No-one has posted the correct result for [imath]\binom{20}{3}[/imath] yet (both post#2 and #5 contain mistakes). I'm sure that @Loki123 can get this right now, with another attempt, since @JeffM probably just mistyped something in the actual calculation.

Yes. I divided 18 by 6 and came up with 1. Maybe I should stop posting today. I seem to be making arithmetic mistakes all over.

 

[Steven G]

Yes. I divided 18 by 6 and came up with 1. Maybe I should stop posting today. I seem to be making arithmetic mistakes all over.

Sitting in the corner for a spell does wonders when it comes to not making silly arithmetic mistakes.
I do have to admit that saying that 18/6=1 is a good one!

 

[Steven G]

I have been given a hint that the answer is not smaller than 1000 and not bigger than 7000. With my way I would get waaaay bigger numbers... Where am I going wrong?? View attachment 32308

smaller than 1000 and not bigger than 7000 equates to smaller than 1000--for the record.

 

[Cubist]

smaller than 1000 and not bigger than 7000 equates to smaller than 1000--for the record.

But the OP is, "the answer is not smaller than 1000 and not bigger than 7000". Therefore if "w" is the number of ways then 1000 ≤ w ≤ 7000

In programming I avoid boolean variable names that begin with the prefix "not" since this often causes me a lot of confusion For example:-

boolean not_Steven_is_in_the_corner = false;

 

[pka]

You're assuming that the numbers can't be repeated. Where does it say that?

The problem clearly states that three numbers are to be chosen. If a number is repeated the would not be three.
The sum of three odd integers is an odd integer. The sum of two even integers and an odd integer is an odd integer.
The set of integers [imath]\{1,2,3,\cdots,38,39,40\}[/imath] contains twenty odds and twenty evens.
There are [imath]\binom{20}{3}=1140[/imath] to choose three from twenty. See This
How many ways are there to choose two evens and one odd?
[imath][/imath][imath][/imath]

 

[Steven G]

But the OP is, "the answer is not smaller than 1000 and not bigger than 7000". Therefore if "w" is the number of ways then 1000 ≤ w ≤ 7000

In programming I avoid boolean variable names that begin with the prefix "not" since this often causes me a lot of confusion For example:-

boolean not_Steven_is_in_the_corner = false;

I knew my corner time was coming up soon. You got me good.

 

[Loki123]

This is what I got