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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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 1 green and 1 blue dye is:
Solution
Total dyes = $5 + 4 + 3 = 12$. Total combinations (excluding none) = $2^{12} - 1 = 4095$. Exclude sets with no green or no blue: - No green → choose from $(4+3)=7$: $2^7 - 1 = 127$. - No blue → choose from $(5+3)=8$: $2^8 - 1 = 255$. Add back those with no green and no blue → only red $(3)$: $2^3 - 1 = 7$. Hence required = $4095 - (127 + 255 - 7) = 4095 - 375 = 3720$.Online Test Series, Information About Examination,
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