
How many sets of five marbles include either the lavender one or exactly one yellow one but not both colors if a bag contains three red marbles, four green ones, one lavender, five yellows, and five orange marbles?
Answer
539.7k+ views
Hint: Here, we have to use the concept of combination to find the number of possible ways. The combination is used when we have to choose various ways in which objects from a set may be selected. So, we will form the combination equation for selecting the one marble from the six marbles of color lavender and yellow. Then we will multiply it with the combination equation of selecting 4 marbles from the remaining marbles in the bag to get the required answer.
Complete step by step solution:
It is given that a bag contains three red marbles, four green ones, one lavender, five yellows, and five orange marbles.
We have to select the sets of five marbles including either the lavender one or exactly one yellow one but not both colors.
Firstly we will form the combination equation of selecting only 1 marble from the total of yellow and lavender marbles which are equal to 6 marbles as there are only one lavender, five yellow marbles are available. Therefore, the equations forms as
\[{ ^6}{C_1} = \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}}\]
\[{ \Rightarrow ^6}{C_1} = \dfrac{{6!}}{{5!}} = 6\]……………………. \[\left( 1 \right)\]
Now we will form the combination equation of selecting the 4 marbles from the remaining marbles in the bag which is equal to 12 which includes three red marbles, four green ones, and five orange marbles. Therefore, the equation forms as
\[{ \Rightarrow ^{12}}{C_4} = \dfrac{{12!}}{{\left( {12 - 4} \right)!}}\]
\[{ \Rightarrow ^{12}}{C_4} = \dfrac{{12!}}{{4!8!}} = \dfrac{{12 \times 11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}} = 495\]……………………. \[\left( 2 \right)\]
Now we will multiply the equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] to get the number of possible sets of five marbles including either the lavender one or exactly one yellow one but not both colors. Therefore, we get
Numbers of possible sets of five marbles include either the lavender one or exactly one yellow one but not both colors \[ = 6 \times 495 = 2970\]
Hence the numbers of possible sets of five marbles include either the lavender one or exactly one yellow one but not both colors is equal to 2970.
Note:
Permutations may be defined as the different ways in which a collection of items can be arranged. For example: The different ways in which the numbers 1, 2 and 3 can be grouped together, taken all at a time, are \[123,{\rm{ }}132,{\rm{ }}213{\rm{ }}231,{\rm{ }}312,{\rm{ }}321\].
So, Number of permutations of n things, taken r at a time, denoted by \[^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Combinations may be defined as the various ways in which objects from a set may be selected. For example: The different selections possible from the numbers 1, 2, 3 taking 2 at a time, are \[{\rm{12, 23 and 31}}\]
So, Number of combinations possible from n group of items, taken r at a time, denoted by \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].
Complete step by step solution:
It is given that a bag contains three red marbles, four green ones, one lavender, five yellows, and five orange marbles.
We have to select the sets of five marbles including either the lavender one or exactly one yellow one but not both colors.
Firstly we will form the combination equation of selecting only 1 marble from the total of yellow and lavender marbles which are equal to 6 marbles as there are only one lavender, five yellow marbles are available. Therefore, the equations forms as
\[{ ^6}{C_1} = \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}}\]
\[{ \Rightarrow ^6}{C_1} = \dfrac{{6!}}{{5!}} = 6\]……………………. \[\left( 1 \right)\]
Now we will form the combination equation of selecting the 4 marbles from the remaining marbles in the bag which is equal to 12 which includes three red marbles, four green ones, and five orange marbles. Therefore, the equation forms as
\[{ \Rightarrow ^{12}}{C_4} = \dfrac{{12!}}{{\left( {12 - 4} \right)!}}\]
\[{ \Rightarrow ^{12}}{C_4} = \dfrac{{12!}}{{4!8!}} = \dfrac{{12 \times 11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}} = 495\]……………………. \[\left( 2 \right)\]
Now we will multiply the equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] to get the number of possible sets of five marbles including either the lavender one or exactly one yellow one but not both colors. Therefore, we get
Numbers of possible sets of five marbles include either the lavender one or exactly one yellow one but not both colors \[ = 6 \times 495 = 2970\]
Hence the numbers of possible sets of five marbles include either the lavender one or exactly one yellow one but not both colors is equal to 2970.
Note:
Permutations may be defined as the different ways in which a collection of items can be arranged. For example: The different ways in which the numbers 1, 2 and 3 can be grouped together, taken all at a time, are \[123,{\rm{ }}132,{\rm{ }}213{\rm{ }}231,{\rm{ }}312,{\rm{ }}321\].
So, Number of permutations of n things, taken r at a time, denoted by \[^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Combinations may be defined as the various ways in which objects from a set may be selected. For example: The different selections possible from the numbers 1, 2, 3 taking 2 at a time, are \[{\rm{12, 23 and 31}}\]
So, Number of combinations possible from n group of items, taken r at a time, denoted by \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].
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