A bag has 5 red balls and 4 green balls. In how many ways can you make

Question

A bag has 5 red balls and 4 green balls. In how many ways can you make a selection if you have to take at least 1 red ball and 1 green ball?

A. 12
B. 17
C. 20
D. 345
E. 465

GMATinsight · Answer

5 red balls and 4 green balls

Every ball can be selected or deselected in 2 ways

Total ways of making selection of red balls from 0 to 5 balls = 2*2*2*2*2

But we need at least 1 red balls so ways of selecting at least one red ball out of 5 = 2^5 - 1 = 31

Total ways of making selection of Green balls from 0 to 4 balls = 2*2*2*2

But we need at least 1 red balls so ways of selecting at least one red ball out of 4 = 2^4 - 1 = 15

Total Favourable Cases = 31*15 = 465

Answer: Option E

CareerGeek · Answer

Hi
2^4 - 1 = 15. You have mistakenly written as 17.
Pls correct

Posted from my mobile device

nick1816 · Answer

Total possible selections = 2^{9}

Number of selections in which no green ball is selected= 2^5

Number of selections in which no red ball is selected= 2^4

Number of ways you make a selection if you have to take at least 1 red ball and 1 green ball = 2^9-2^5-2^4+1 = 465

{We have to add 1 because we subtract the case 2 times in which none of the balls is selected}

Dillesh4096  it should be mentioned in the question that the balls of same color are distinct or similar.

preetamsaha · Answer

A bag has 5 red balls and 4 green balls.
we have to take at least 1 red ball and 1 green ball.
no of combinations :
(red , green )= (1,1), (1,2), (1,3), (1,4) ; 
 (2,1), (2,2), (2,3), (2,4) ;
 (3,1), (3,2), (3,3), (3,4) ; 
 (4,1), (4,2), (4,3), (4,4) ;
 (5,1), (5,2), (5,3), (5,4) . 
so, total no of cases : 5C1 *  +5C2 *  +5C3 *  + 5C4 *  +5C5 *  
 = 5*  + 10*  + 10*  + 5*  + 1*  
 =   *  
 = 31*15
 = 465

nC0 + nC1 + nC2 + ... + nCn = 2^n  in case the numbers of balls are in large quantities ; 
here, 4C1+4C2+4C3+4C4 = 2^4 - 1 = 15
 5C1+5C2+5C3+5C4+ 5C5 = 2^5 -1 = 31 ]

cristianov · Answer

Cases
1 red ball combines with green balls
(1;1) (1;2) (1;3) (1;4)
2 red ball combines with green balls
(2;1) (2;2) (2;3) (2;4)
3 red ball combines with green balls
(3;1) (3;2) (3;3) (3;4)
4 red ball combines with green balls
(4;1) (4;2) (4;3) (4;4)
5 red ball combines with green balls
(5;1) (5;2) (5;3) (5;4)

20 cases in total

preetamsaha · Answer

cristianov

1 red ball combines with green balls
(1;1) (1;2) (1;3) (1;4)
2 red ball combines with green balls
(2;1) (2;2) (2;3) (2;4)
3 red ball combines with green balls
(3;1) (3;2) (3;3) (3;4)
4 red ball combines with green balls
(4;1) (4;2) (4;3) (4;4)
5 red ball combines with green balls
(5;1) (5;2) (5;3) (5;4)
 these are the probable combinations . out these all combinations we have to find total no of cases . for example (1,1) ; for this combination , we can find one red ball out of 5 in 5C1 ways  AND  one green ball in 4C1 ways . so we  MULTIPLY BOTH  .

cristianov · Answer

The question says "ways" not "cases", therefore the number of combination should be calculated

preetamsaha · Answer

cristianov  yes , that is why  nick1816  has mentioned all the balls are distinct and it should explicitly be mentioned in question. so,  nick1816  do u think if its not mentioned the answer will be 20 ?

nick1816 · Answer

If all the balls are distinct, it must be 465. 
If 5 red balls are similar and 4 green balls are similar, it should be 20.

bumpbot · Answer

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A bag has 5 red balls and 4 green balls. In how many ways can you make

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