Binomial coefficient problem

W

wefsefef

New member
Joined
Dec 6, 2021
Messages
1
renderTimingPixel.png

In a supermarket, there are five different kinds of candy bags to choose from, but the supermarket has only six bags left of three of the kinds as well as seven bags of the last two kinds. Show that you can choose ten candy bags in 866 ways.
Explanation
We can select seven or more of a particular kind by first selecting seven of a kind and then choose three bags freely among all five kinds. This can be done in 3+5-1 choose 5-1 ways = 35 ways. You can choose 8 or more of a particular kind by first selecting 8 of the kind and there after 2 bags among all five kinds, which is 2+5-1 choose 5-1 ways=15 ways. Therefore there are 1001-3*35-2*15=866 combinations
I understand the explanation. What I don't get if I try to visualize it. Then I have five boxes which i will seperate by L.
L L L L
Then you can do as an example 8L
and then you have from before 15 combinations for example 8L1L0L1L0
But cant you do this with 8 being in every box like
For example at first 8 is in the first box
But you can also do it where L is in the second box like 0L8L and so on
And then it shuld be 5*15 ways. Or what am I doing wrong? Thanks!
 
Look at the problem. You have 1001 total ways which I suppose calculated by (1+x+x^2+x^3…x^10)^5. That means each item has qty of 10 and there are 5 items total. According to the condition 3 bags has max qty of 6, and that implies you CANNOT choose 7 or more for these 3, and that’s calculated based on minus 3*choosing 7 of a particular bag then others (3 bags left) from freely choosing among 5. And you CANNOT choose 8 or more for 2 bags that has a max qty of 7. Hence minus 2*choosing 8 of a particular bag then others (2 bags left) from freely choosing among 5. That would be my understanding.
 
2+5-1 choose 5-1 implies 2 bags choosing freely from 5 so **||||, that 8 bags doesn’t go anywhere in that formula.
 
You must log in or register to reply here.
Top