Lucy-Lastic Posted October 2, 2011 Share Posted October 2, 2011 This is this weeks Y3 homework: 'Tim has nine cards, each with a different number from 1 to 9 on it. He put the cards in three piles so that the total of each pile was 15. Can you find all the different ways Tim could have done this?' Junior and I have found 7 combinations so far before he got bored - just wondering how many there really are (what is the formula for working this one out - combinations and permutations was not one of my strong points in maths)? I have enjoyed doing this one much more than Junior LOL (but he has got some lovely number bond practise done ) Link to comment Share on other sites More sharing options...
Eater Sundae Posted October 2, 2011 Share Posted October 2, 2011 I think there are 2 ways: 1, 5, 9 + 2, 6, 7 + 3, 4, 8 and 1, 6, 8 + 2, 4, 9 + 3, 5, 7 Edit. I prepared a detailed answer, explaining how I did it, but it disappeared into the ether as I posted. I will close everything down, and if it still exists I'll post it later. Edit - just re-read your post, and I realise I have misunderstood the puzzle. I will try again. Link to comment Share on other sites More sharing options...
Lucy-Lastic Posted October 2, 2011 Author Share Posted October 2, 2011 You don't need 3 in each pile though, just 3 piles with each pile adding up to 15. I have got 7 ways so far. 9+6, 8+7, 5+4+3+2+1 9+5+1, 8+4+3, 7+6+2 9+4+2, 8+6+1, 3+7+5 9+3+2+1, 6+5+4, 8+7 8+5+2, 4+1+3+7, 9+6 6+4+3+2, 9+5+1, 8+7 6+5+3+1, 8+7, 9+4+2 Link to comment Share on other sites More sharing options...
Jeffrey Shaw Posted October 2, 2011 Share Posted October 2, 2011 The cards numbers total 45. Each pile's numbers total 15. So you need as many '15' totals as possible. The pile with card 1 lacks 14 more. The 14 could include 2 (so lacks 12 more) or 3-but-not-2 (so lacks 11 more) And so on... Link to comment Share on other sites More sharing options...
Eater Sundae Posted October 2, 2011 Share Posted October 2, 2011 Right, another try. If I start with "9". This must be in a pile with cards that total "6", which can be 6 OR 5+1 OR 4+2 or 3+2+1. (The "9" can never be in the same pile as an "8" or a "7"). If I consider each combination in turn, and then look at the next piles which can go with it, starting with "8". This can be 8+7 or 8+ other numbers (depending on which are available, ie are not already in the "9" pile) Following this, I come up with the following: 9+6 & 8+7 & 5+4+3+2+1 9+6 & 8+5+2 & 7+4+3+1 9+6 & 8+4+3 & 7+5+2+1 9+6 & 8+4+2+1 & 7+5+3 9+5+1 & 8+7 & 6+4+3+2 9+5+1 & 8+4+3 & 7+6+2 9+4+2 & 8+7 & 6+5+3+1 9+4+2 & 8+6+1 & 7+5+3 9+3+2+1 & 8+7 & 6+5+4 So, I've found 9 ways. Link to comment Share on other sites More sharing options...
Jeffrey Shaw Posted October 2, 2011 Share Posted October 2, 2011 Yes- that is the same approach (mirror image) as in my post. Link to comment Share on other sites More sharing options...
chem1st Posted October 2, 2011 Share Posted October 2, 2011 edit: you beat me to it, think you've got them all now Link to comment Share on other sites More sharing options...
Eater Sundae Posted October 2, 2011 Share Posted October 2, 2011 The next job is the other part of the OP, Is there a formula that explains/calculates the number of answers? Anybody got any ideas? Link to comment Share on other sites More sharing options...
Eater Sundae Posted October 2, 2011 Share Posted October 2, 2011 Also, what age is Y3? Link to comment Share on other sites More sharing options...
Lucy-Lastic Posted October 2, 2011 Author Share Posted October 2, 2011 Junior is 7 so it took him quite a while to work through the ones that we had as his mental arithmatic is not brilliant:) Thanks for the extra 2. The question doesn't ask for the formula, that is just me that wonders how you can work it it. I seem to remember needing to use factorials and the like when working out combinations. I know you guys like a little challenge every now and agin Link to comment Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.