MCQ
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.
$[$Hint: Possible numbers of choosing or not choosing $5$ green dyes$, 4$ blue dyes and $3$ red dyes are $2^5, 2^4$ and $2^3 ,$ respectively$.]$
  • A
    $3600$
  • $3720$
  • C
    $3800$
  • D
    $3600$

Answer

Correct option: B.
$3720$
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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