Question:

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 atleast one green and one blue dye is

Updated On: Jul 6, 2022
  • 3600
  • 3720
  • 3800
  • 3500
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The Correct Option is B

Solution and Explanation

At. least one green dye can be selected out of $5$ green dyes in $2^{5}-1$, i.e. in $31$ ways. Similarly at least one blue dye can be selected out of $4$ in $2^{4}-1$ in $15$ ways. For red dyes there is no restriction; you may include it or not include it. Thus there are two ways of disposing of each of red dye. Thus the total number of selection of red dye is $2^{3}$ $=8$ Hence the required number of ways $31 \times 15 \times 8=3720$.
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Concepts Used:

Permutations and Combinations

Permutation:

Permutation is the method or the act of arranging members of a set into an order or a sequence. 

  • In the process of rearranging the numbers, subsets of sets are created to determine all possible arrangement sequences of a single data point. 
  • A permutation is used in many events of daily life. It is used for a list of data where the data order matters.

Combination:

Combination is the method of forming subsets by selecting data from a larger set in a way that the selection order does not matter.

  • Combination refers to the combination of about n things taken k at a time without any repetition.
  • The combination is used for a group of data where the order of data does not matter.