The Hardest Maths Questions You'll Ever Try

  • Thread starter Doomotron
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No, these questions from a maths textbook that I'm asking aren't the hardest you'll ever try but I'll (semi) admit defeat and ask for help from GTP's denizens.

Firstly:

How many different symbols will be needed so that at least 1 million different three-symbol arrangements can be made?

And secondly:

Four students are going to be chosen, from a group of 10, to represent their school at a conference. Only one of Aziz, Baz or Chris is to be included. In how many ways can the students be chosen?
 
HenrySwanson
How many different symbols will be needed so that at least 1 million different three-symbol arrangements can be made?
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100? 100×100×100 = 1,000,000

I think. :lol:

HenrySwanson
Four students are going to be chosen, from a group of 10, to represent their school at a conference. Only one of Aziz, Baz or Chris is to be included. In how many ways can the students be chosen?
Click to expand...
Not enough information. If, however unlikely, there are three named Aziz, three Baz and four Chris (or some other combination of 3/3/4), it can't be done.
 
TB
100? 100×100×100 = 1,000,000

I think. :lol:
Click to expand...
That's what I got - the book gives an answer of 1000 :confused:

So far through 16 chapters all the answers I've come across have been correct (or at least I've agreed with them) and I'm stumped as to how it could be 1000

TB
Not enough information. If, however unlikely, there are three named Aziz, three Baz and four Chris (or some other combination of 3/3/4), it can't be done.
Click to expand...
Hmmm I see what you mean....

I got 140 as my answer as I assumed there was only 1 Aziz, Baz and Chris

I started with 7!/4!X3! = 35
which would be the combinations of 4 students picked with none of them being Aziz, Baz or Chris.

Then I did 8!/4!X4! = 70
70-35 = 35

To get a combination of Aziz + 3 other classmates

Then 35 X3 = 105
105 +35 = 140

To find all possible combinations, i.e. Aziz + 3 others, Baz with 3 others, Chris with 3 others, and 4 students without Aziz, Baz or Chris.

The answer however is not 140....
 
HenrySwanson
And secondly:

Four students are going to be chosen, from a group of 10, to represent their school at a conference. Only one of Aziz, Baz or Chris is to be included. In how many ways can the students be chosen?
Click to expand...

Is the answer provided by the book 105, or something else?
 
HenrySwanson
Yes!

How did you reach that?
Click to expand...

From this:

Four students are going to be chosen, from a group of 10, to represent their school at a conference. Only one of Aziz, Baz or Chris is to be included. In how many ways can the students be chosen?
Click to expand...

Assumptions:
- Group of 10 includes Aziz (A), Baz (B) or Chris (C)
- 4 students are to be chosen
- Only one of A, B or C is to be included. (I interpreted this to mean one of A, B or C must be included, but only one of them)

Therefore, there are 10-3=7 other students that can be chosen

In this case, order doesn't matter (whether order is A, Student 1, Student 2, Student 3 or Student 2, Student 3, Student 1, A, or something else, it's still the same group of people), so this is a combinations problem.

4 students are to be chosen. 1 of which must be A, B, or C, therefore, 3 students are to be chosen from the group of 7.

7 Choose 3 yields 35, 7!/(3!*(7-3)!)=35

Last slot goes to A, B or C, and there are 3 possibilities, so 3*35=105 combinations.
 
Crash
From this:



Assumptions:
- Group of 10 includes Aziz (A), Baz (B) or Chris (C)
- 4 students are to be chosen
- Only one of A, B or C is to be included. (I interpreted this to mean one of A, B or C must be included, but only one of them)

Therefore, there are 10-3=7 other students that can be chosen

In this case, order doesn't matter (whether order is A, Student 1, Student 2, Student 3 or Student 2, Student 3, Student 1, A, or something else, it's still the same group of people), so this is a combinations problem.

4 students are to be chosen. 1 of which must be A, B, or C, therefore, 3 students are to be chosen from the group of 7.

7 Choose 3 yields 35, 7!/(3!*(7-3)!)=35

Last slot goes to A, B or C, and there are 3 possibilities, so 3*35=105 combinations.
Click to expand...
Ahh that's the only way I could get it to be 105 too (3X35), but I assumed there could be cases without A, B or C and so added an extra 35.

Thanks, it seems that it's a poorly worded question which happens to follow a question with a wrong answer (the symbol one)
 

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