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In a packet there are rn different books, n different pens and p different pencils. The number of selections of atleast one article of each type from the packet is
- a)2m+n+p - 1
- b)(m+n)(n+1)(p+1)-1
- c)2m+n+p
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
In a packet there are rn different books, n different pens and p diffe...
The required number of ways
= Total number of ways of selecting any number of books from m different books, any number of pens from n different, pens and any number of pencils from p different pencils - 1.
= Total number of ways of selecting any number of books from m different books, any number of pens from n different, pens and any number of pencils from p different pencils - 1.
Most Upvoted Answer
In a packet there are rn different books, n different pens and p diffe...
The required number of ways
= Total number of ways of selecting any number of books from m different books, any number of pens from n different, pens and any number of pencils from p different pencils - 1.
= Total number of ways of selecting any number of books from m different books, any number of pens from n different, pens and any number of pencils from p different pencils - 1.
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In a packet there are rn different books, n different pens and p diffe...
Understanding the Problem
To solve the problem, we need to determine the number of ways to select at least one article of each type from a packet containing rn different books, n different pens, and p different pencils.
Types of Articles
- Books: rn different options
- Pens: n different options
- Pencils: p different options
Calculating Selections
1. Books Selection:
- Each book can either be selected or not, leading to 2 choices (select or not) for each book. Therefore, for rn books, the total combinations are 2^(rn). However, we need at least one book, so we subtract the case where no book is selected:
- Total selections for books = 2^(rn) - 1.
2. Pens Selection:
- Similar to books, for n pens, the total combinations are 2^n. Subtracting the case where no pen is selected gives:
- Total selections for pens = 2^n - 1.
3. Pencils Selection:
- For p pencils, the same logic applies, yielding:
- Total selections for pencils = 2^p - 1.
Combining Selections
Now, we combine these selections. The total number of ways to select at least one article from each type is the product of the selections from each category:
- Total selections = (2^(rn) - 1)(2^n - 1)(2^p - 1).
However, the question asks for the total selections of at least one article of each type, which simplifies to:
- Total selections = 2^(rn + n + p) - 1.
Thus, the correct answer is option (a) 2^(rn+n+p) - 1.
To solve the problem, we need to determine the number of ways to select at least one article of each type from a packet containing rn different books, n different pens, and p different pencils.
Types of Articles
- Books: rn different options
- Pens: n different options
- Pencils: p different options
Calculating Selections
1. Books Selection:
- Each book can either be selected or not, leading to 2 choices (select or not) for each book. Therefore, for rn books, the total combinations are 2^(rn). However, we need at least one book, so we subtract the case where no book is selected:
- Total selections for books = 2^(rn) - 1.
2. Pens Selection:
- Similar to books, for n pens, the total combinations are 2^n. Subtracting the case where no pen is selected gives:
- Total selections for pens = 2^n - 1.
3. Pencils Selection:
- For p pencils, the same logic applies, yielding:
- Total selections for pencils = 2^p - 1.
Combining Selections
Now, we combine these selections. The total number of ways to select at least one article from each type is the product of the selections from each category:
- Total selections = (2^(rn) - 1)(2^n - 1)(2^p - 1).
However, the question asks for the total selections of at least one article of each type, which simplifies to:
- Total selections = 2^(rn + n + p) - 1.
Thus, the correct answer is option (a) 2^(rn+n+p) - 1.
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