
From 3 cocoa nuts, 4 apples, and 2 oranges, how many selections of fruit can be made, taking at least one of each kind?
Answer
460.2k+ views
Hint: 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.
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 \[1\] apple from \[4\] apples is \[^4{C_1}\] .The number of ways of selecting \[1\] cocoa nut from \[3\] cocoa nut is \[^3{C_1}\] .The number of ways of selecting \[1\] oranges from \[2\] oranges is \[^2{C_1}\] .
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,
\[{ \Rightarrow ^4}{C_1}{ \times ^3}{C_1}{ \times ^2}{C_1}\]
That gives us,
\[ \Rightarrow 4 \times 3 \times 2 \times 1\]
Hence, we get
\[ \Rightarrow 24\]
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 \[24\].
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 \[^n{P_r}\] and \[^n{C_r}\] respectively, and are defined as,
\[{ \Rightarrow ^n}{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
And
\[{ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] .
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.
| Apples | Cocoa nuts | Oranges |
| 4 | 3 | 2 |
| 3 | 2 | 1 |
| 2 | 1 | |
| 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 \[1\] apple from \[4\] apples is \[^4{C_1}\] .The number of ways of selecting \[1\] cocoa nut from \[3\] cocoa nut is \[^3{C_1}\] .The number of ways of selecting \[1\] oranges from \[2\] oranges is \[^2{C_1}\] .
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,
\[{ \Rightarrow ^4}{C_1}{ \times ^3}{C_1}{ \times ^2}{C_1}\]
That gives us,
\[ \Rightarrow 4 \times 3 \times 2 \times 1\]
Hence, we get
\[ \Rightarrow 24\]
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 \[24\].
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 \[^n{P_r}\] and \[^n{C_r}\] respectively, and are defined as,
\[{ \Rightarrow ^n}{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
And
\[{ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] .
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