Question

There are 5 English, 4 Sanskrit and 3 Telugu books. Two books from each group are to be arranged on a shelf. The number of possible arrangements is:
 $ \begin{align}
  & \left( a \right)\left( 180 \right)6! \\
 & \left( b \right)\left( 12 \right)7! \\
 & \left( c \right)7! \\
 & \left( d \right)180 \\
\end{align} $

Answer 1

Hint: First find a solution possible from each group. Now find the number of arrangements possible for selected groups. Look at definitions of sum rule, product rule. Try to select one of them to apply here. Now find the total number of arrangements.

Complete step-by-step answer:
Given collection of books in the question contains the type:
 5 English books, 4 Sanskrit books, 3 Telugu books. We have to select 2 books from each group from above. Selection is analogous method to combination which is defined as:
Combinations: It is a mathematical technique that determines the number of possible arrangements in a collection of items where the other of selection does not matter in combination; you can select items in any order. Formulation is given by $ ^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} $
Now we will divide our selections into 3 cases as:
Case (1): Selecting 2 books out of the 5 English books. By applying the definition of combination, we get it as: English books = $ ^{5}{{C}_{2}}.......\left( 1 \right) $
Case (2): Selecting 2 books out of the 4 Sanskrit books. By applying the definition of combinations, we get it as: Sanskrit books = $ ^{4}{{C}_{2}}..........\left( 2 \right) $
Case (3): Selecting 2 books out of the 3 Telugu books. By applying the definition of combinations, we get it as: Telugu books = $ ^{3}{{C}_{2}}..........\left( 3 \right) $
Now we have 6 books in our hands which we need to arrange in a shelf.
Number of ways to arrange n different things are $ \left( n! \right) $ ways. By substituting the value of n to be 6, we get the total ways: Total ways = $ 6!..........\left( 4 \right) $
Rule of sum: In combinatorics, the rule of sum or addition principle is basic counting principle; it is simply defined as, if there are A ways of doing P work B ways of doing Q work. Given P, Q works cannot be done together. Total number of ways to do both P, Q are given by $ \left( A+B \right) $ ways.
Rule of Product: In combinatorics, the rule of product or multiplication principle is basic counting principle. It is simply defined as, if there are A ways of doing P work and B ways of doing Q work. Given P, Q works can be done at a time. Total number of ways to do both P, Q works are given by $ \left( A.B \right) $ ways.
From definitions of product rule, sum rule we select product rule. By applying product rule to equation $ \left( 1 \right),\left( 2 \right),\left( 3 \right), $ we get them as: Total selections = $ \left( English \right)\times \left( Sanskrit \right)\times \left( Telugu \right) $
By substituting their values, we get the equation in form of: Total selections = $ ^{5}{{C}_{2}}{{\times }^{4}}{{C}_{2}}{{\times }^{3}}{{C}_{2}}. $
By expanding their values by combination definition, we get it as:
Total selection = $ \dfrac{5!}{3!2!}\times \dfrac{4!}{2!2!}\times \dfrac{3!}{2!1!}=\dfrac{5\times 4}{2}\times \dfrac{4\times 3}{2}\times 3 $ .
As we know total ways for each selection Total arrangements:
Total arrangements = $ \left( Total\text{ Ways} \right)\times \left( Total\text{ Selections} \right) $ .
By substituting their values, we get the equation as:
Total arrangements = $ \left( 6! \right)\times \left( \dfrac{20}{2}\times \dfrac{12}{2}\times 3 \right) $
By simplifying we can write it in the form of: Total arrangements = $ 180\times \left( 6! \right) $
Therefore the option $ \left( a \right) $ is the correct answer for this question.

Note: Be careful while calculating each subject selection as it is the base for solutions. If not given anything generally we assume them as different books. If they are identical then the whole solution changes. The arrangements as fractional must also be done carefully.