## mathmath333 one year ago An urn has 5 red balls ,4 black balls,3 blue balls,2 yellow balls and 1 green ball. In how many ways can 5 balls be selected.

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{An urn has 5 red balls ,4 black balls,3 blue balls,2 yellow balls}\hspace{.33em}\\~\\ & \normalsize \text{and 1 green ball.}\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can 5 balls be selected.}\hspace{.33em}\\~\\ \end{align}}

2. mathmath333

I dnt think this is stars ans bas problem

3. mathmath333

stars and bars are used to find integer solutions of the equation eg. a+b+c=x

4. imqwerty

well there are no other conditions given like there shuld be atleast 2 yellow ones or something like that so we can consider like u have a total of 15balls nd u need to find out number of ways to select 5balls so it wuld be 15*14*13*11*10 but wait we can have cases like this-|dw:1442129915453:dw| so we are getting similar cases :) nd we need to fix them but how :)

5. anonymous

you can write out explicitly the options r = red, b = black , B = blue, y = yellow, g = green r1, r2, r3 , r4, r5 b1, b2, b3 , b4, B1, B2, B3 y1, y2 g there are 5+4+3+2+1 = 15 balls there are 15*14*13*12*11 ways to choose 5 balls. I am assuming order counts because you can select r1,r2,r3,r4,r5 in different ways

6. imqwerty

yea sry that ws not 15*14*13*11*10 i forgot the 12 it wuld be 15*14*13*12*11

7. anonymous

the question is a bit ambiguous . is it asking how many 'different' ways of selecting 5 balls there are

8. anonymous

and we have indistinguishable balls

9. mathmath333

hey same colored balls are identical

10. mathmath333

answer given is 71

11. anonymous

thanks, i will work with this

12. mathmath333

15*14*13*12*11 this works when all 15 balls are distinct

13. imqwerty

yea :)

14. anonymous

did you get 71 imquert?

15. imqwerty

nope m still wrkin on it :)

16. mathmath333

i m frutated by this que from 2 days

17. imqwerty

:D lets finish off the frustration today :)

18. imqwerty

hehe see my mod powers :) - http://prntscr.com/8frqsl

19. mathmath333

it happns to every body

20. anonymous

another way to look at this is to list the possible number of colored balls Red = 0,1,2,3,4,5 Black = 0,1,2,3,4 Blue = 0,1,2,3 yellow = 0,1,2 green = 0,1 how many ways can you add up to 5?

21. anonymous

starting with Red = 5, we have 5 + 0 + 0 + 0 + 0 4 + 1 + 0 + 0 + 0 4 + 0 + 1 + 0 + 0 4 + 0 + 0 + 1 + 0 4 + 0 + 0 + 0 + 1

22. anonymous

then 3 + 2 + 0 + 0 + 0 3 + 0 + 2 + 0 + 0 3 + 0 + 0 + 2 + 0 then list with 3 reds. This is brute force method and long.

23. anonymous

'5 + 0 + 0 + 0 + 0 ' means 5 red + 0 black + 0 blue + 0 yellow + 0 green

24. imqwerty

:D nice question btw

25. mathmath333

Their should be some trick in this question as it is to be solved in 1-2 min

26. anonymous

there are 7 ways to add up to 5 The seven partitions of 5 are: 5 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1 now count how many ways you can fit these in the boxes, with the restriction that first ball is between 0 and 5 second ball is between 0 and 4, etc

27. mathmath333

is 7 the answer

28. mathmate

The answer is 71 by enumeration, but I have not found a systematical way to do it. If you want to see the list, let me know.

29. mathmate

Here's the list of the combinations, first number is the green [0,1], last number is red [0,5]. Perhaps someone can figure out the logic. (0, 0, 0, 0, 5) (0, 0, 0, 1, 4) (0, 0, 0, 2, 3) (0, 0, 0, 3, 2) (0, 0, 0, 4, 1) (0, 0, 1, 0, 4) (0, 0, 1, 1, 3) (0, 0, 1, 2, 2) (0, 0, 1, 3, 1) (0, 0, 1, 4, 0) (0, 0, 2, 0, 3) (0, 0, 2, 1, 2) (0, 0, 2, 2, 1) (0, 0, 2, 3, 0) (0, 0, 3, 0, 2) (0, 0, 3, 1, 1) (0, 0, 3, 2, 0) (0, 1, 0, 0, 4) (0, 1, 0, 1, 3) (0, 1, 0, 2, 2) (0, 1, 0, 3, 1) (0, 1, 0, 4, 0) (0, 1, 1, 0, 3) (0, 1, 1, 1, 2) (0, 1, 1, 2, 1) (0, 1, 1, 3, 0) (0, 1, 2, 0, 2) (0, 1, 2, 1, 1) (0, 1, 2, 2, 0) (0, 1, 3, 0, 1) (0, 1, 3, 1, 0) (0, 2, 0, 0, 3) (0, 2, 0, 1, 2) (0, 2, 0, 2, 1) (0, 2, 0, 3, 0) (0, 2, 1, 0, 2) (0, 2, 1, 1, 1) (0, 2, 1, 2, 0) (0, 2, 2, 0, 1) (0, 2, 2, 1, 0) (0, 2, 3, 0, 0) (1, 0, 0, 0, 4) (1, 0, 0, 1, 3) (1, 0, 0, 2, 2) (1, 0, 0, 3, 1) (1, 0, 0, 4, 0) (1, 0, 1, 0, 3) (1, 0, 1, 1, 2) (1, 0, 1, 2, 1) (1, 0, 1, 3, 0) (1, 0, 2, 0, 2) (1, 0, 2, 1, 1) (1, 0, 2, 2, 0) (1, 0, 3, 0, 1) (1, 0, 3, 1, 0) (1, 1, 0, 0, 3) (1, 1, 0, 1, 2) (1, 1, 0, 2, 1) (1, 1, 0, 3, 0) (1, 1, 1, 0, 2) (1, 1, 1, 1, 1) (1, 1, 1, 2, 0) (1, 1, 2, 0, 1) (1, 1, 2, 1, 0) (1, 1, 3, 0, 0) (1, 2, 0, 0, 2) (1, 2, 0, 1, 1) (1, 2, 0, 2, 0) (1, 2, 1, 0, 1) (1, 2, 1, 1, 0) (1, 2, 2, 0, 0) ('count=', 71)

30. mathmath333

am i amazed ,how did u use the brute force

31. mathmate

When I cannot solve a problem, brute force is required, and the computer is my best friend for that purpose! lol If you use Python, I can send you the code.

32. mathmath333

ok

33. mathmate

Table of number of ways to obtain total no. of balls progressively sum of number of balls 0 1 2 3 4 5 red (5) 1 1 1 1 1 1 black(4) 1 1 1 1 1 (5+4) 1 2 3 4 5 5 blue(3) 1 1 1 1 (5+4+3) 1 3 6 10 14 17 yellow(2) 1 1 1 (5...2) 1 4 10 19 30 41 green(1) 1 1 (5...1) 1 5 14 29 49 71 Number of ways to make the given sum (0 - 5) for given colour combinations. Explanations: The first line of the table enumerates the ways to pick the red balls to get a given total of one to five balls. The next line is similar, but for black balls. The third line is the number of ways to choose x balls (0<=x<=5) when red and black are combined. After that, we add progressively blue, yellow and green. The last line is when there are all 15 balls, with 71 ways to get a total of five balls. Following are some examples: for 5 reds and 4 blacks, there are 1*1=1 way to get a sum of 0 ball 1*1+1*1=2 ways to get 1 ball, ... 1*1+1*1+1*1+1*1+1*1=5 ways to get 4 balls 1*1+1*1+1*1+1*1+1*1=5 ways to get 5 balls The result is shown on the line (5+4) For 5 reds, 4 blacks, and 3 blues, tag on the blues below the line (5+4) and work similarly. 1*1=1 way to get a sum of zero ball. 2*1+1*1=3 ways to get 1 ball, 3*1+2*1+1*1=6 ways to get 2 balls, ... 3*1+4*1+5*1+5*1=17 ways to get a sum of 5 balls. Continuing this way, we have, with all 15 balls, 71 ways to get a sum of exactly 5 balls.