IIT JEE , AIEEE Forum - Mathematics , Algebra

vishal16

1) given 5 different dyes- 4 diff blue dyes and 3 diff red dyes-- the no of combination of dyes which can be chosen taking at least 1 green and 1 blue dye is

a) 3600 b) 3720 c)3800 d) none of thes

2) there r 3 piles of identical red, blue and green balls and each pile contains atleast 10 balls--the no of ways of selecting 10 balls if twice as many red balls as green balls r to be selected.

a) 3 b) 4 c) 6 d)8

sameerh522

1) 3720

rajatsen91

1. 3720
The number of ways are (2^5-1)(2^4-1)(2^3)=3720

jatinroxx

Answer to second question : 3

There'r just 3 cases possible...

'coz if one is twice the other,dat wud account for 3n balls and remaining will be identical...

Therefore no. of possible ways = [10/3] = 3

[] represents modular division...........

chinky85

thanx but sorry
i dont understand wat u r saying jatin roxx

jatinroxx

It goes like this...

If one green ball is selected, two red balls must be selected, but no more red or green can be taken as the condition would not be met then...

Next u take 2 green balls and then 3...

When u take 4 green, the total number of balls wud exceed 12 and that is wat is taken care of by []. It gives the whole number of the division...

avinash.sharma

Answer to second question : 3

The possible cases are

Green Balls	Red Balls	Blue Balls
1	2	7
2	4	4
3	6	1

metal

The first answer is indeed 3720 and the second answer is indeed 3

thechamp

the answer is "4" for 2nd question

3 cases have been defined by the others
the 4th case is
blue : 10 balls
green : 00 balls
red : 2 x 00 = 00 balls

sravanbrahma

ans : c

ibmbasith

3

venkat_tatolu

2)Answer is 4.
Let the no. of green balls be x.Then the no. of red balls is2x.Let the no. of blue balls be y.Then,
x+2x+y=10 Implies 3x+y=10 Implies y=10-3x.
Clearly,x can take values 0,1,2,3.The corresponding values of y are
10,7,4 and 1.Thus the possibilities are (0,10,0),(2,7,1),(4,4,2) and
(6,1,3),where (r,b,g) denotes the no. of red,blue and green balls.

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