## mathmath333 one year ago Counting question

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{Their are 4 apples, 5 mangoes and 6 watermelons .}\hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways }\hspace{.33em}\\~\\ & \normalsize \text{i.) I can purchase at least each of them. }\hspace{.33em}\\~\\ & \normalsize \text{ii.) I can purchase at least one of them. }\hspace{.33em}\\~\\ \end{align}}

2. ganeshie8

notice that apples have 5 states : {0, 1, 2, 3, 4} mangoes have 6 states : {0, 1, 2, 3, 4, 5} watermelons have 7 states : {0, 1, 2, 3, 4, 5, 6}

3. mathmath333

yes

4. mathmath333

is this correct =(2^4-1)*(2^5-1)*(2^6-1)

5. anonymous

I believe so

6. anonymous

the english is a bit unclear to me " I can purchase at least each of them"

7. mathmath333

but it is given 4*5*6 ways

8. mathmath333

in book

9. anonymous

thats the answer for part a) ?

10. mathmath333

yes

11. ganeshie8

for partb, is the answer 5*6*7 - 1 ?

12. mathmath333

yes for part b

13. ganeshie8

it seems textbook is treating all the fruit types as indistinguishable @jayzdd

14. anonymous

yes , hmm i guess i treated them as distinguishable

15. mathmath333

all apples are are identical and so all mango

16. anonymous

For the apple choice you can pick 1 apple 2 apples 3 apples 4 apples a total of 4 ways to select the apples Then multiply this by the number of selections you can make for mango 1 mango 2 mangoes 3 mangoes 4 mongoes 5 mangoes a total of 5 choices and then similarly for the watermelon there are 6 ways to select them by multiplication principle, 4*5*6

17. mathmath333

ok

18. anonymous

for part b) now include the zero case

19. anonymous

For the apple choice you can pick 0 apples, 1 apple, 2 apples, 3 apples , 4 apples a total of 5 ways to select the apples Then multiply this by the number of selections you can make for mango 0 mango 1 mango 2 mangoes 3 mangoes 4 mangoes 5 mangoes a total of 6 ways to select And then similarly for the watermelon 0 watermelon, 1 watermelon, ... 6 watermelon there are 7 ways to select them By multiplication principle, 5*6*7 But the directions say at least one fruit, so we cannot have 0 apple and 0 mangoe, and 0 watermelon. so we subtract 1

20. mathmath333

like these method in the previuos question there were 5 methods for bulb so how can i judge distinguishable bulbs or fruits

21. anonymous

The bulbs were fixed by the socket, so there was an ordering to them. lightbulb 1, lightbulb2, ... lightbulb 5 Here there is no natural order to the fruits, since they are indistinguishable. If they want to treat the fruits as distinguishable, the directions would explicitly say it. otherwise assume indistinguishable

22. anonymous

|dw:1440349712575:dw|

23. anonymous

the sockets are distinguishable i would say , and there are 2 states for each socket (on or off)

24. anonymous

|dw:1440349885112:dw|

25. anonymous

If the lightbulbs had no sockets, then the lightbulbs would be indistinguishable. But then there would be no light either. Since lightbulbs had light, we could assume they were connected to distinguishable sockets, and cannot move. This forces an ordering to the lightbulbs that make them distinguishable.

26. anonymous

The lightbulbs connected to the sockets are fixed in space and bolted down, because they are constructed that way by the carpenter. But the apples and mangoe fruits are not bolted down, so what matters is how many you pick, not which one

27. anonymous

Think of it this way. When you select the fruits, imagine you are blindfolded. There are three bags, a bag for apples , a bag for mangoes, bag for watermelons. You are instructed to select as many fruits from each bag as you want. The apples are all the same (let's imagine for the sake of argument). The mangoes are all exactly the same and the watermelons the same. you can select 0 apples, 1 apple, 2 apples... up to 4 apples. you can select 0 mangoes up to 5 mangoes, etc Doing this blindfold experiment would not make sense for the lighbulb example.

28. mathmath333

sry was afk