
From \[12\] books , the difference between the number of ways a selection of \[5\] books when one specified book is always excluded and one specified book is always included , is equal to?
Answer
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Hint: We are given \[12\] books and we have to find the difference between the number of ways a section of \[5\] books when one specified book is always excluded and one specified book is always included. For this type of question we will use the concept of combinations to get the desired result. If one book is always included, then we have to choose only four books. Similarly, if we have one book that is always excluded, then we have one less option to choose from the total available options. So, we will apply the combinations and permutations formula accordingly.
Complete step-by-step solution:
Permutation relates to the act of arranging all the members of a set into some sequence or order. Permutations occur, in more or less prominent ways, in almost every area of mathematics. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. In smaller cases, it is possible to count the number of combinations
A permutation is the choice of \[r\] things from a set of \[n\] things without replacement and where the order matters.
\[{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}\]
A combination is the choice of \[r\] things from a set of \[n\] things without replacement and where order doesn't matter.
\[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
When one specified book is always included, then \[4\] out of \[11\]will be chosen in \[{}^{11}{C_4} = \dfrac{{11!}}{{4!(11 - 4)!}} = 330\]
When one specified book is always included, then \[5\] out of \[11\] will be chosen in \[{}^{11}{C_5} = \dfrac{{11!}}{{5!(11 - 5)!}} = 462\]
Hence, the difference between the number of ways a section of \[5\] books when one specified book is always excluded and one specified book is always included will be \[ = 462 - 330 = 132\]
Note: Wisely decide which method should be used permutation or combination according to the statement in the question. We must remember the combination and permutation formula so as to solve the question and get to the correct answer. We must remember to take the mod value when calculating the difference of both the answers obtained separately.
Complete step-by-step solution:
Permutation relates to the act of arranging all the members of a set into some sequence or order. Permutations occur, in more or less prominent ways, in almost every area of mathematics. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. In smaller cases, it is possible to count the number of combinations
A permutation is the choice of \[r\] things from a set of \[n\] things without replacement and where the order matters.
\[{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}\]
A combination is the choice of \[r\] things from a set of \[n\] things without replacement and where order doesn't matter.
\[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
When one specified book is always included, then \[4\] out of \[11\]will be chosen in \[{}^{11}{C_4} = \dfrac{{11!}}{{4!(11 - 4)!}} = 330\]
When one specified book is always included, then \[5\] out of \[11\] will be chosen in \[{}^{11}{C_5} = \dfrac{{11!}}{{5!(11 - 5)!}} = 462\]
Hence, the difference between the number of ways a section of \[5\] books when one specified book is always excluded and one specified book is always included will be \[ = 462 - 330 = 132\]
Note: Wisely decide which method should be used permutation or combination according to the statement in the question. We must remember the combination and permutation formula so as to solve the question and get to the correct answer. We must remember to take the mod value when calculating the difference of both the answers obtained separately.
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