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Question: On the occasion of the Deepawali festival, each student of a class sends greeting cards to the other...

On the occasion of the Deepawali festival, each student of a class sends greeting cards to the others. If there are 20 students in the class, then the total number of getting cards exchanged by the students is
(a) 20C3\left( a \right){{\text{ }}^{20}}{{C}_{3}}
(b)2.20C2\left( b \right)\text{2}{{\text{.}}^{20}}{{C}_{2}}
(c)2.20P2\left( c \right)\text{2}{{\text{.}}^{20}}{{P}_{2}}
(d) None of these

Explanation

Solution

To solve this question, we will first calculate the number of cards that each student is giving, by understanding the data given in the question. Finally, we will calculate the number of cards with 20 students as the product of 20 and 19. Then, we will check which option gives the same value by using the formula of nPr^{n}{{P}_{r}} and nCr^{n}{{C}_{r}} is given by nPr=n!(n−r)!^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} and nCr=n!r!(n−r)!.^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.

Complete step-by-step solution
We are given that there are 20 students in a class. There is an exchange of greeting cards with the other students. Basically, one student shares greeting cards with all the rest of the students. And this process is done by each of 20 students. Now, because one student gets a card from all other students except himself and there are 20 students. Each student gets a card from 19 students. Because each student shares 19 cards and there are a total of 20 students. Therefore the cards shared by 20 students is
⇒20×19=380\Rightarrow 20\times 19=380
Therefore, 380 cards were shared in all. Now, we have to match the given option.
Consider option (a). 20C3^{20}{{C}_{3}}
Using the formula of combination given as nCr=n!r!(n−r)!.^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}. So, we have,
20C3=20!(20−3)!(3)!^{20}{{C}_{3}}=\dfrac{20!}{\left( 20-3 \right)!\left( 3 \right)!}
⇒20C3=20×19×18×17!17!3×2×1{{\Rightarrow }^{20}}{{C}_{3}}=\dfrac{20\times 19\times 18\times 17!}{17!3\times 2\times 1}
⇒20C3=20×19×3≠20×19{{\Rightarrow }^{20}}{{C}_{3}}=20\times 19\times 3\ne 20\times 19
So, option (a) is the wrong answer, as 20×19×3≠20×19.20\times 19\times 3\ne 20\times 19.
Now, consider option (b). 2.20C2{{2.}^{20}}{{C}_{2}}
Using the formula, nCr=n!r!(n−r)!.^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}. We get,
⇒2.20C2=2×20!(20−2)!2!\Rightarrow {{2.}^{20}}{{C}_{2}}=2\times \dfrac{20!}{\left( 20-2 \right)!2!}
⇒2.20C2=2×20×19×18!18!×2×1\Rightarrow {{2.}^{20}}{{C}_{2}}=2\times \dfrac{20\times 19\times 18!}{18!\times 2\times 1}
⇒2.20C2=20×19×18!18!\Rightarrow {{2.}^{20}}{{C}_{2}}=\dfrac{20\times 19\times 18!}{18!}
⇒2.20C2=20×19\Rightarrow {{2.}^{20}}{{C}_{2}}=20\times 19
As, 20×19=20×19,20\times 19=20\times 19, hence, option (b) is the right answer.
Now, let us consider the option (c). 2.20P2.{{2.}^{20}}{{P}_{2}}.
Using the formula, nPr=n!(n−r)!^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} We have,
2.20P2=2×20!(20−2)!{{2.}^{20}}{{P}_{2}}=2\times \dfrac{20!}{\left( 20-2 \right)!}
⇒2.20P2=2×20×19×18!18!\Rightarrow {{2.}^{20}}{{P}_{2}}=2\times \dfrac{20\times 19\times 18!}{18!}
⇒2.20P2=2×20×19\Rightarrow {{2.}^{20}}{{P}_{2}}=2\times 20\times 19
And as 20×19≠2×20×1920\times 19\ne 2\times 20\times 19 so option (c) is the wrong answer.
Hence, the total number of cards exchanged by the student is 20×19=2.20C2.20\times 19={{2.}^{20}}{{C}_{2}}.. Hence, option (b) is the right answer.

Note: We observe the options here in these are of the type nCr, nPr.^{n}{{C}_{r}},{{\text{ }}^{n}}{{P}_{r}}. Also, 20C2^{20}{{C}_{2}} and 20P2^{20}{{P}_{2}} type which is nothing but the selecting and arranging 2 from 20 given values respectively. Hence, this kind of selection or arranging is not possible in our question. So, the approach that must be followed to solve this question is by using the value 20×19.20\times 19. There are chances of calculation mistakes while finding the value of each option, so be careful not to interchange the formulas and choose the wrong option.