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Permutation and Combinations — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSPermutation and Combinations1 Mark
MCQ
Choose the correct answer.
Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is.
✓

Answer

  1. 3720

Solution:

At least one green dye can be chosen in $^5\text{C}_1+\ ^5\text{C}_2\ +\ ^5\text{C}_3+\ ^5\text{C}_4+\ ^5\text{C}_5=22^5-1$ways.

At least one blue dye can be chosen in $^4\text{C}_1+\ ^4\text{C}_2\ +\ ^4\text{C}_3+\ ^4\text{C}_4=2^4-1$ ways.

Any number of red dyes can be chosen in $^3\text{C}_0+\ ^3\text{C}_1\ +\ ^3\text{C}_2+\ ^3\text{C}_3=2^3$ ways.

so, total nnumber of required selection $=(2^5-1)\times(2^4-1)\times2^3=3720$

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