Question

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.

Answer 1

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

Answer 2

I dnt think this is stars ans bas problem

Answer 3

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

Answer 4

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-

so we are getting similar cases :) nd we need to fix them but how :)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

and we have indistinguishable balls

Answer 9

hey same colored balls are identical

Answer 10

answer given is 71

Answer 11

thanks, i will work with this

Answer 12

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

Answer 13

yea :)

Answer 14

did you get 71 imquert?

Answer 15

nope m still wrkin on it :)

Answer 16

i m frutated by this que from 2 days

Answer 17

:D lets finish off the frustration today :)

Answer 18

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

Answer 19

it happns to every body

Answer 20

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?

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

:D nice question btw

Answer 25

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

Answer 26

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

Answer 27

is 7 the answer

Answer 28

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.

Answer 29

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)

Answer 30

am i amazed ,how did u use the brute force

Answer 31

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.

Answer 32

ok

Answer 33

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.