
To deal with winters, Mr. A wears at least 3 T-shirts every day. He has a total of 4 T-shirts. Find the number of days he can deal with winter, without wearing the same arrangement on more than one day.
Answer
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Hint: According to the question, he will not wear the same arrangement more than one day. So, we have to find a total number of arrangements which is possible by taking at least 3 T-shirts out of available 4 T-shirts. So, find the total number of arrangements possible by taking 3 T-shirts from available 4 T-shirts or by taking all the 4 available T-shirts.
Complete step-by-step answer:
According to the question, the total available T-shirts are 4 and he will wear at least 3 T-shirts daily and he won’t repeat any arrangement.
So, the total number of days he can deal with the winter will be equal to the total number of possible arrangements with the given constraint.
He wears at least 3 T-shirts daily. So, we have two options: either he will wear 3 T-shirts or 4 T-shirts.
Total arrangements will be the sum of these two possible options.
Let us first find the total number of arrangements if he will wear 3 T-shirts.
Total number of arrangements if he will wear 3 T-shirts = Total number of ways of selecting 3 T-shirts out of 4 T-shirts $\times $ Arrangement of the selected 3 T-shirts
We know that total number of ways of selection of ‘r’ objects out of n-distinct objects in $^{n}{{C}_{r}}$.
And total number of arrangement of ‘n’ objects on ‘n’ places $=n!$
So, we can write
Total number of ways of selecting 3 T-shirts out of 4 T-shirts = $^{4}{{C}_{3}}$
Arrangement of the selected 3 T-shirts = 3!
Total number of arrangements if he will wear 3 T-shirts = $^{4}{{C}_{3}}\times 3!$
Similarly,
Total number of arrangements if he will wear 4 T-shirts = Total number of ways of selecting 4 T-shirts out of 4 T-shirts $\times $ Arrangement of the selected 4 T-shirts
So, we can write
Total number of ways of selecting 4 T-shirts out of 4 T-shirts = $^{4}{{C}_{4}}$
Arrangement of the selected 4 T-shirts = 4!
Total number of arrangements if he will wear 4 T-shirts = $^{4}{{C}_{4}}\times 4!$
Now, we will get the
Total number of arrangements = Total number of arrangements if he will wear 3 T-shirts + Total number of arrangements if he will wear 4 T-shirts
$\Rightarrow \left( ^{4}{{C}_{3}}\times 3! \right)+\left( ^{4}{{C}_{4}}\times 4! \right)$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\ \text{ and }n!=n\times \left( n-1 \right)\times .......\times 1$
So, we can write
Total number of arrangements
\[\begin{align}
& =\left( \dfrac{4!}{3!\times 1!}\times 3! \right)+\left( \dfrac{4!}{4!\ 0!}\times 4! \right) \\
& \Rightarrow 4!+4! \\
& \Rightarrow 2\times 4! \\
& \Rightarrow 2\times \left( 4\times 3\times 2\times 1 \right) \\
& \Rightarrow 48 \\
\end{align}\]
So, the total number of arrangements under given constraints will be 48.
Hence, the total number of days he can deal with this winter will be 48.
Note: Students can make mistakes by not considering the cases when they will wear 4 T-shirts on a day. In question it is given that he wears at least 3 T-shirts on a day, it means he can wear greater than or equal to 3 T-shirts on a day and he has a total 4 T-shirts, so he can wear maximum 4 T-shirts in a day. To get the total number of arrangements, students must add the cases and not multiply them. This mistake must be avoided.
Complete step-by-step answer:
According to the question, the total available T-shirts are 4 and he will wear at least 3 T-shirts daily and he won’t repeat any arrangement.
So, the total number of days he can deal with the winter will be equal to the total number of possible arrangements with the given constraint.
He wears at least 3 T-shirts daily. So, we have two options: either he will wear 3 T-shirts or 4 T-shirts.
Total arrangements will be the sum of these two possible options.
Let us first find the total number of arrangements if he will wear 3 T-shirts.
Total number of arrangements if he will wear 3 T-shirts = Total number of ways of selecting 3 T-shirts out of 4 T-shirts $\times $ Arrangement of the selected 3 T-shirts
We know that total number of ways of selection of ‘r’ objects out of n-distinct objects in $^{n}{{C}_{r}}$.
And total number of arrangement of ‘n’ objects on ‘n’ places $=n!$
So, we can write
Total number of ways of selecting 3 T-shirts out of 4 T-shirts = $^{4}{{C}_{3}}$
Arrangement of the selected 3 T-shirts = 3!
Total number of arrangements if he will wear 3 T-shirts = $^{4}{{C}_{3}}\times 3!$
Similarly,
Total number of arrangements if he will wear 4 T-shirts = Total number of ways of selecting 4 T-shirts out of 4 T-shirts $\times $ Arrangement of the selected 4 T-shirts
So, we can write
Total number of ways of selecting 4 T-shirts out of 4 T-shirts = $^{4}{{C}_{4}}$
Arrangement of the selected 4 T-shirts = 4!
Total number of arrangements if he will wear 4 T-shirts = $^{4}{{C}_{4}}\times 4!$
Now, we will get the
Total number of arrangements = Total number of arrangements if he will wear 3 T-shirts + Total number of arrangements if he will wear 4 T-shirts
$\Rightarrow \left( ^{4}{{C}_{3}}\times 3! \right)+\left( ^{4}{{C}_{4}}\times 4! \right)$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\ \text{ and }n!=n\times \left( n-1 \right)\times .......\times 1$
So, we can write
Total number of arrangements
\[\begin{align}
& =\left( \dfrac{4!}{3!\times 1!}\times 3! \right)+\left( \dfrac{4!}{4!\ 0!}\times 4! \right) \\
& \Rightarrow 4!+4! \\
& \Rightarrow 2\times 4! \\
& \Rightarrow 2\times \left( 4\times 3\times 2\times 1 \right) \\
& \Rightarrow 48 \\
\end{align}\]
So, the total number of arrangements under given constraints will be 48.
Hence, the total number of days he can deal with this winter will be 48.
Note: Students can make mistakes by not considering the cases when they will wear 4 T-shirts on a day. In question it is given that he wears at least 3 T-shirts on a day, it means he can wear greater than or equal to 3 T-shirts on a day and he has a total 4 T-shirts, so he can wear maximum 4 T-shirts in a day. To get the total number of arrangements, students must add the cases and not multiply them. This mistake must be avoided.
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