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Question
Answer
How many squares are there on a standard chessboard?
64
How many white squares are there on a standard chessboard?
32
How many symmetries does a regular pentagon have?
10
How many edges does a cuboid have?
12
How many different scores are possible when you roll a pair of fair dice?
11
How many prime numbers are there between 1 and 10?
4
How many factors does 120 have?
16
How many months have 28 days?
12
Question
Answer
How many seconds are there in a week?
604800
A mystery bag contains lots of red and blue beans. You draw beans one at a time from the bag while blindfolded. What is the minimum number of beans you need to draw to guarantee you will get two beans of the same colour?
3
Four people race each other in the 100m. How many possible outcomes does the race have? (Assume ties/draws are impossible.)
24
64 men enter a singles tennis tournament. How many matches must be played to determine the winner?
63
How many different tickets are possible in a lottery where 3 numbers are drawn from the numbers 1 to 8? (The order of the drawn numbers does not matter.)
56
Into how many regions do three non-parallel lines divide the plane? (The three lines do not all pass through a common point.)
7
What is the minimum number of randomly chosen people you need to have at a party to guarantee that either three of them are mutual friends or three of them are mutual strangers?
Thanks, I have edited the question so hopefully everything is right now.
I was meaning to ask for the number of possible tickets (where ordering doesn't matter) rather than the number of possible outcomes (where it sounds like ordering does matter).
My thinking was as follows: When three numbers are drawn, say {a,b,c}, the order in which they are drawn doesn't matter, so the {a,b,c} ticket is considered the same as the tickets {a,c,b}, {b,a,c}, {b,c,a}, {c,a,b}, {c,b,a}. Therefore you have to divide 8x7x6 by the number of possible orderings which is 3x2x1 = 6, hence number of possible tickets = 8x7.
what does knowing about basketball have to do with counting? All the other questions involve some sort of numerical reasoning. That one is just a numerical fact.
The diagram at 2:00 in the linked video (here: https://www.youtube.com/watch?v=H6TCPlVP1bI) shows why 5 people is not enough and the explanation starting at 2:34 shows that 6 people is always enough.
= 8! / (5! * 3!)
= 5! * 6 * 7 * 8 / (5! * 3!)
= 6 * 7 * 8 / 3!
= 6 * 7 * 8 / 6
= 7 * 8
= 56
I was meaning to ask for the number of possible tickets (where ordering doesn't matter) rather than the number of possible outcomes (where it sounds like ordering does matter).
My thinking was as follows: When three numbers are drawn, say {a,b,c}, the order in which they are drawn doesn't matter, so the {a,b,c} ticket is considered the same as the tickets {a,c,b}, {b,a,c}, {b,c,a}, {c,a,b}, {c,b,a}. Therefore you have to divide 8x7x6 by the number of possible orderings which is 3x2x1 = 6, hence number of possible tickets = 8x7.