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Question: From 3 cocoa nuts, 4 apples, and 2 oranges, how many selections of fruit can be made, taking at leas...

From 3 cocoa nuts, 4 apples, and 2 oranges, how many selections of fruit can be made, taking at least one of each kind?

Explanation

Solution

In the above given question, we are given a number of fruits such as 3 cocoa nuts, 4 apples, and 2 oranges. We have to find the number of selections that can be made by taking at least one fruit of each kind at a time. In order to approach the solution, we have to use the method of permutations and combinations to find the possible number of required ways.

Complete step by step answer:
Given that, we have 3 cocoa nuts, 4 apples, and 2 oranges. We have to find the number of ways of selecting at least one fruit of each kind. That means the possible number of each fruit can be taken as represented in the table below.

ApplesCocoa nutsOranges
432
321
21
1

The table shows the possible combinations of the number of fruits that can be taken.Now we have to calculate the required number of ways.So, the number of ways of selecting 11 apple from 44 apples is 4C1^4{C_1} .The number of ways of selecting 11 cocoa nut from 33 cocoa nut is 3C1^3{C_1} .The number of ways of selecting 11 oranges from 22 oranges is 2C1^2{C_1} .
Therefore, the number of ways of selections that can be made by taking at least one fruit of each kind at a time is given by,
4C1×3C1×2C1{ \Rightarrow ^4}{C_1}{ \times ^3}{C_1}{ \times ^2}{C_1}
That gives us,
4×3×2×1\Rightarrow 4 \times 3 \times 2 \times 1
Hence, we get
24\Rightarrow 24
That is the required number of ways.

Therefore, the number of ways of selection that can be made by taking at least one fruit of each kind at a time is 2424.

Note: The permutations and combinations are defined as the various ways in which objects from a number of sets may be selected, generally without replacement to form new subsets. This selection of subsets is called a permutation when the order of selection is a factor, whereas it is called a combination when the order is not a factor. They are denoted by nPr^n{P_r} and nCr^n{C_r} respectively, and are defined as,
nPr=n!(nr)!{ \Rightarrow ^n}{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}
And
nCr=n!r!(nr)!{ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} .