In how many ways 2 different prizes are awarded to 15 students, without giving both to the same student?
Answer
586.2k+ views
Hint: To obtain the number of ways 2 different prizes can be awarded to 15 students such that both awards don’t go to the same student we will use combination technique. Firstly we will find ways in which first prize can be given to the student. Then we will find that from the remaining student who can get the second prize. Finally we will multiply both the possibility and get the answer.
Complete step by step answer:
So we have to find ways in which 2 prizes are distributed among 15 students.
So for ${{1}^{st}}$ prize we have 15 students who can get it so the combination will be as follows:
${}^{15}{{C}_{1}}$….$\left( 1 \right)$
We know that
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$….$\left( 2 \right)$
Using above formula in equation (1) we get,
$\begin{align}
& {}^{15}{{C}_{1}}=\dfrac{15!}{1!\left( 15-1 \right)!} \\
& {}^{15}{{C}_{1}}=\dfrac{15\times 14!}{14!} \\
\end{align}$
${}^{15}{{C}_{1}}=15$…..$\left( 3 \right)$
So the first prize can be given in 15 ways.
Next as we know one student already got the first prize so the student left for the second prize is 14.
So 14 students can get the second prize in the below ways:
${}^{14}{{C}_{1}}$….$\left( 4 \right)$
Using formula (2) in equation (4) we get,
$\begin{align}
& {}^{14}{{C}_{1}}=\dfrac{14!}{1!\left( 14-1 \right)!} \\
& {}^{14}{{C}_{1}}=\dfrac{14\times 13!}{13!} \\
\end{align}$
${}^{14}{{C}_{1}}=14$……$\left( 5 \right)$
So we can give second prize to 14 students in 14 ways.
For finding the total ways we will multiply equation (3) and (5) as below:
Total ways $=15\times 14$
Total ways $=210$
Hence in 210 ways 2 different prizes can be awarded to 15 students without giving both to the same student.
Note: Combination is used to find the number of possible arrangements in a collection of items where the order in which the selection is done doesn’t matter. It is different from permutation as in permutation the order in which the items are selected matters.
Complete step by step answer:
So we have to find ways in which 2 prizes are distributed among 15 students.
So for ${{1}^{st}}$ prize we have 15 students who can get it so the combination will be as follows:
${}^{15}{{C}_{1}}$….$\left( 1 \right)$
We know that
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$….$\left( 2 \right)$
Using above formula in equation (1) we get,
$\begin{align}
& {}^{15}{{C}_{1}}=\dfrac{15!}{1!\left( 15-1 \right)!} \\
& {}^{15}{{C}_{1}}=\dfrac{15\times 14!}{14!} \\
\end{align}$
${}^{15}{{C}_{1}}=15$…..$\left( 3 \right)$
So the first prize can be given in 15 ways.
Next as we know one student already got the first prize so the student left for the second prize is 14.
So 14 students can get the second prize in the below ways:
${}^{14}{{C}_{1}}$….$\left( 4 \right)$
Using formula (2) in equation (4) we get,
$\begin{align}
& {}^{14}{{C}_{1}}=\dfrac{14!}{1!\left( 14-1 \right)!} \\
& {}^{14}{{C}_{1}}=\dfrac{14\times 13!}{13!} \\
\end{align}$
${}^{14}{{C}_{1}}=14$……$\left( 5 \right)$
So we can give second prize to 14 students in 14 ways.
For finding the total ways we will multiply equation (3) and (5) as below:
Total ways $=15\times 14$
Total ways $=210$
Hence in 210 ways 2 different prizes can be awarded to 15 students without giving both to the same student.
Note: Combination is used to find the number of possible arrangements in a collection of items where the order in which the selection is done doesn’t matter. It is different from permutation as in permutation the order in which the items are selected matters.
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