Answer :
Certainly! Here is a detailed, step-by-step solution for the given question.
Let's break it down into each part:
1. The first and last positions will be occupied by Geography books:
We have a total of [tex]\( 5 \)[/tex] Geography books (G1, G2, G3, G4, G5) and [tex]\( 6 \)[/tex] History books (H1, H2, H3, H4, H5, H6), making a total of [tex]\( 11 \)[/tex] books.
- Total possible arrangements:
There are [tex]\( 11! \)[/tex] ways to arrange [tex]\( 11 \)[/tex] distinct books.
- Arrangements where first and last are Geography books:
- Choose 2 out of the 5 Geography books for the first and last positions: [tex]\( \binom{5}{2} \times 2! = 10 \times 2 = 20 \)[/tex] ways (since we can select and arrange 2 books in 20 ways).
- The remaining 9 books (3 Geography + 6 History) can be arranged in [tex]\( 9! \)[/tex] ways.
Therefore, the number of favorable arrangements is:
[tex]\[ 20 \times 9! \][/tex]
- Probability:
[tex]\[ P(\text{first and last are Geography}) = \frac{20 \times 9!}{11!} = \frac{20 \times 9!}{11 \times 10 \times 9!} = \frac{20}{110} = \frac{2}{11} \][/tex]
2. All the Geography books will be grouped together:
Consider the 5 Geography books as a single unit. Now we have 7 units in total (the single Geography unit + 6 History books).
- Total arrangements of these 7 units:
[tex]\( 7! \)[/tex] ways.
- Within the single Geography unit, the 5 Geography books can be arranged among themselves in [tex]\( 5! \)[/tex] ways.
- Therefore, the number of favorable arrangements is:
[tex]\[ 7! \times 5! \][/tex]
- Total possible arrangements of all 11 books:
[tex]\( 11! \)[/tex]
- Probability:
[tex]\[ P(\text{Geography books together}) = \frac{7! \times 5!}{11!} \][/tex]
Simplifying,
[tex]\[ P(\text{Geography books together}) = \frac{7! \times 5!}{11 \times 10 \times 9 \times 8 \times 7!} = \frac{5!}{11 \times 10 \times 9 \times 8} = \frac{120}{7920} = \frac{1}{66} \][/tex]
3. All the Geography books and all the History books will be grouped together:
Now consider Geography books as one unit and History books as another unit.
- Total number of ways to arrange these 2 units is [tex]\( 2! = 2 \)[/tex] ways.
- The 5 Geography books among themselves can be arranged in [tex]\( 5! \)[/tex] ways.
- The 6 History books can be arranged among themselves in [tex]\( 6! \)[/tex] ways.
Therefore, the number of favorable arrangements is:
[tex]\[ 2! \times 5! \times 6! \][/tex]
- Total possible arrangements of all 11 books is:
[tex]\( 11! \)[/tex]
- Probability:
[tex]\[ P(\text{Geography books and History books together}) = \frac{2 \times 5! \times 6!}{11!} \][/tex]
Simplifying,
[tex]\[ P(\text{Geography books and History books together}) = \frac{2 \times 120 \times 720}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5!} = \frac{2 \times 120 \times 720}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 120} = \frac{2}{11 \times 10 \times 9 \times 8 \times 7 \times 6} = \frac{2}{332640} = \frac{1}{166320} \][/tex]
4. All the Geography books will be first, followed by all the History books:
- The 5 Geography books can be arranged in [tex]\( 5! \)[/tex] ways.
- The 6 History books can be arranged in [tex]\( 6! \)[/tex] ways.
Therefore, the number of favorable arrangements is:
[tex]\[ 5! \times 6! \][/tex]
- Total possible arrangements of all 11 books is:
[tex]\( 11! \)[/tex]
- Probability:
[tex]\[ P(\text{All Geography first, followed by all History}) = \frac{5! \times 6!}{11!} \][/tex]
Simplifying,
[tex]\[ P(\text{All Geography first, followed by all History}) = \frac{120 \times 720}{11 \times 10 \times 9 \times 8 \times 7 \times 6!} = \frac{86400}{39916800} = \frac{1}{462} \][/tex]
5. Three boys and four girls are seated randomly on a bench:
- Total arrangements:
There are [tex]\( 7! \)[/tex] ways to arrange [tex]\( 7 \)[/tex] people (3 boys + 4 girls).
The problem does not specify a condition for seating, so the number of favorable outcomes equals the total possible arrangements:
[tex]\[ P(\text{Any seating arrangement}) = 7! \][/tex]
Therefore:
1. Probability the first and last are Geography books : [tex]\( \frac{2}{11} \)[/tex].
2. Probability all Geography books are grouped together : [tex]\( \frac{1}{66} \)[/tex].
3. Probability all Geography books and all History books are grouped together : [tex]\(\frac{1}{166320} \)[/tex].
4. Probability all Geography books are first, followed by all History books : [tex]\( \frac{1}{462} \)[/tex].
5. Total number of seating arrangements for three boys and four girls : [tex]\( 7! = 5040 \)[/tex].
Let's break it down into each part:
1. The first and last positions will be occupied by Geography books:
We have a total of [tex]\( 5 \)[/tex] Geography books (G1, G2, G3, G4, G5) and [tex]\( 6 \)[/tex] History books (H1, H2, H3, H4, H5, H6), making a total of [tex]\( 11 \)[/tex] books.
- Total possible arrangements:
There are [tex]\( 11! \)[/tex] ways to arrange [tex]\( 11 \)[/tex] distinct books.
- Arrangements where first and last are Geography books:
- Choose 2 out of the 5 Geography books for the first and last positions: [tex]\( \binom{5}{2} \times 2! = 10 \times 2 = 20 \)[/tex] ways (since we can select and arrange 2 books in 20 ways).
- The remaining 9 books (3 Geography + 6 History) can be arranged in [tex]\( 9! \)[/tex] ways.
Therefore, the number of favorable arrangements is:
[tex]\[ 20 \times 9! \][/tex]
- Probability:
[tex]\[ P(\text{first and last are Geography}) = \frac{20 \times 9!}{11!} = \frac{20 \times 9!}{11 \times 10 \times 9!} = \frac{20}{110} = \frac{2}{11} \][/tex]
2. All the Geography books will be grouped together:
Consider the 5 Geography books as a single unit. Now we have 7 units in total (the single Geography unit + 6 History books).
- Total arrangements of these 7 units:
[tex]\( 7! \)[/tex] ways.
- Within the single Geography unit, the 5 Geography books can be arranged among themselves in [tex]\( 5! \)[/tex] ways.
- Therefore, the number of favorable arrangements is:
[tex]\[ 7! \times 5! \][/tex]
- Total possible arrangements of all 11 books:
[tex]\( 11! \)[/tex]
- Probability:
[tex]\[ P(\text{Geography books together}) = \frac{7! \times 5!}{11!} \][/tex]
Simplifying,
[tex]\[ P(\text{Geography books together}) = \frac{7! \times 5!}{11 \times 10 \times 9 \times 8 \times 7!} = \frac{5!}{11 \times 10 \times 9 \times 8} = \frac{120}{7920} = \frac{1}{66} \][/tex]
3. All the Geography books and all the History books will be grouped together:
Now consider Geography books as one unit and History books as another unit.
- Total number of ways to arrange these 2 units is [tex]\( 2! = 2 \)[/tex] ways.
- The 5 Geography books among themselves can be arranged in [tex]\( 5! \)[/tex] ways.
- The 6 History books can be arranged among themselves in [tex]\( 6! \)[/tex] ways.
Therefore, the number of favorable arrangements is:
[tex]\[ 2! \times 5! \times 6! \][/tex]
- Total possible arrangements of all 11 books is:
[tex]\( 11! \)[/tex]
- Probability:
[tex]\[ P(\text{Geography books and History books together}) = \frac{2 \times 5! \times 6!}{11!} \][/tex]
Simplifying,
[tex]\[ P(\text{Geography books and History books together}) = \frac{2 \times 120 \times 720}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5!} = \frac{2 \times 120 \times 720}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 120} = \frac{2}{11 \times 10 \times 9 \times 8 \times 7 \times 6} = \frac{2}{332640} = \frac{1}{166320} \][/tex]
4. All the Geography books will be first, followed by all the History books:
- The 5 Geography books can be arranged in [tex]\( 5! \)[/tex] ways.
- The 6 History books can be arranged in [tex]\( 6! \)[/tex] ways.
Therefore, the number of favorable arrangements is:
[tex]\[ 5! \times 6! \][/tex]
- Total possible arrangements of all 11 books is:
[tex]\( 11! \)[/tex]
- Probability:
[tex]\[ P(\text{All Geography first, followed by all History}) = \frac{5! \times 6!}{11!} \][/tex]
Simplifying,
[tex]\[ P(\text{All Geography first, followed by all History}) = \frac{120 \times 720}{11 \times 10 \times 9 \times 8 \times 7 \times 6!} = \frac{86400}{39916800} = \frac{1}{462} \][/tex]
5. Three boys and four girls are seated randomly on a bench:
- Total arrangements:
There are [tex]\( 7! \)[/tex] ways to arrange [tex]\( 7 \)[/tex] people (3 boys + 4 girls).
The problem does not specify a condition for seating, so the number of favorable outcomes equals the total possible arrangements:
[tex]\[ P(\text{Any seating arrangement}) = 7! \][/tex]
Therefore:
1. Probability the first and last are Geography books : [tex]\( \frac{2}{11} \)[/tex].
2. Probability all Geography books are grouped together : [tex]\( \frac{1}{66} \)[/tex].
3. Probability all Geography books and all History books are grouped together : [tex]\(\frac{1}{166320} \)[/tex].
4. Probability all Geography books are first, followed by all History books : [tex]\( \frac{1}{462} \)[/tex].
5. Total number of seating arrangements for three boys and four girls : [tex]\( 7! = 5040 \)[/tex].
