Latest TCS Aptitude Question SOLUTION: Given 5 different green dyes, 4 different blue dyes and 3 different red dyes, the number of combination of dyes taking atleast one green dye and one blue dye, is

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Given 5 different green dyes, 4 different blue dyes and 3 different red dyes, the number of combination of dyes taking atleast one green dye and one blue dye, is

Asked In TCS chaitanya (11 years ago)

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sumit.kumar singh  (11 years ago)

Given 5 different green dyes, 4 different blue dyes and 3 different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is given by
(2^5-1)*(2^4-1)*(2^3)=3720

Is this solution Helpfull? Yes (18) | No (2)  

RAVI KIRAN  (11 years ago)

Answer: The least number of dyes that a combination can have is 2.
(one blue and one green).

Maximum number of dyes that a combination can have is 12
(5G, 4B, 6R).

Atleast one green dye can be selected out of 5 green dyes. The number of ways is



After selecting one or more green dyes, we can select atleast one blue dye out of 4 different blue dyes. The number of ways is


After selecting atleast one green dye and atleast one blue dye, at least one red dye or no red dye can be selected in

= 1 + 3 + 3 + 1 = 8 ways

The total number of combinations (31)(15)(8) = 3720.

Is this solution Helpfull? Yes (13) | No (2)  

shyamlesh dash  (11 years ago)

formula for choosing atleast one n1 and n2 number of things and any number of n3 things is (2^n1 -1)*(2^n2-1)*(2^n3)=31*15*8=3720 ...........thnku :)

Is this solution Helpfull? Yes (13) | No (1)  

chandu161  (11 years ago)

total 12 dyes!!..2 dyes to pick = 12c2 = 66.

atleast 1 green = 5c1 = 5.

atleast 1 blue = 4c1 = 4.

the equation is (5c1 + 4c1)/12c2 = 9/66 = 3/22 ans!!

Is this solution Helpfull? Yes (8) | No (23)  

Sumit Mukhija  (11 years ago)

total number of dyes = 12
thus selecting two dyes of 12 = 12c2 = 66

for selecting one dye of 5 green = 5c1=5
similarly for one dye of 4 blues = 4c1=4

therefore, selecting 1 blue AND 1 green = 4c1 * 5c1
= 4*5
=20
thus, 20/66= 10/33

Is this solution Helpfull? Yes (5) | No (10)  

pavan pavvi  (11 years ago)

no of ways of selectng atleat 1 green dye is
5c1+5c2+5c3+5c4+5c5=31
no of ways of selectng atleast 1 blue dye is 4c1+4c2+4c3+4c4=15

no of ways of selecting 0 or mor red dyes is 1+3+3+1=8
so ewe get 31*15*8=3720

Is this solution Helpfull? Yes (5) | No (1)  

prashanthi  (11 years ago)

selecting one green dye from 5 is 5c1=5
selecting one blue dye from 4 is 4c1=4
selecting 2 dyes from total 12 dyes is 12c2=66
5+4/66=3/22

Is this solution Helpfull? Yes (4) | No (9)  

Gaurav Singh  (11 years ago)

selecting 1 green and 1 blue=5*4=20 combinations
selecting remaining 10 different dyes=2^10=1024 combinations
total possibilities=1024*20=20480 combinations

Is this solution Helpfull? Yes (2) | No (5)  

thejasri balaraju  (11 years ago)

ans is >=9.we have to add the highest of two numbers because for example if we take 7 it may be 4 blue and 3 red or if we take 8 it may be 5 green and 3 red so the min value is 9.

Is this solution Helpfull? Yes (1) | No (6)  

Gaurav Singh  (11 years ago)

selecting 1 green and 1 blue=5*4=20
selecting remaining 10 different dyes=2^10=1024
total possibilities=1044

Is this solution Helpfull? Yes (1) | No (6)  

Sneha paliwal  (11 years ago)

ans: here the combination dyes to select that is 2 or 3 not mentioned.

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disha verma  (11 years ago)

@ pavan plz explain 1+3+3+1

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Poulami Ghosh  (11 years ago)

5*4*4. as because we can take 1-4 no of blue dyes,1-5 nos of green dyes & 0-3 nos of red dyes.

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