Maharashtra Board Class 11 Maths Part 2 Chapter 6 Permutations and Combinations 6.5 Solutions

Question 1. In how many different ways can 8 friends sit around a table?
Answer:
We know that 'n' persons can sit around a table in (n - 1)! ways
- 8 friends can sit around a table in 7! ways
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.
- 8 friends can sit around a table in 5040 ways.
In simple words: To arrange 'n' distinct items in a circle, the number of ways is (n-1)!. For 8 friends, it's (8-1)! = 7!, which calculates to 5040 distinct arrangements.

🎯 Exam Tip: Remember the formula for circular permutations (n-1)! for distinct objects. Common mistakes include using n! instead of (n-1)! for circular arrangements.

Question 2. A party has 20 participants and a host. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Answer:
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 - 1)! = 20! ways.
When two particular participants be seated on either side of the host.
Host takes chair in 1 way.
These 2 persons can sit on either side of host in 2! ways
Once host occupies his chair, it is not circular permutation any more.
Remaining 18 people occupy their chairs in 18! ways.
- Total number of arrangement possible if two particular participants be seated on either side of the host = 2! × 18!
In simple words: For 21 people in a circular arrangement, there are 20! ways. If the host is fixed and two specific people are on either side (which can be arranged in 2! ways), the remaining 18 people can be arranged in 18! ways, leading to 2! × 18! total arrangements.

🎯 Exam Tip: Break down complex problems. First, find total circular arrangements. Then, for conditions involving fixed positions, treat the fixed group as a unit, and consider both internal arrangements within the unit and external arrangements. Remember that fixing one position in a circular permutation converts it to a linear permutation for the remaining items.

Question 3. Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are
(i) always together.
(ii) never together.
Answer:
(i) Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit.
They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total 23) which can be done in (23 - 1)! = 22! ways.
- The total number of arrangements if two specified delegates are always together = 22! × 2!

(ii) When 2 specified delegates are never together then, the other 22 delegates can participate in a round table discussion in (22 - 1)! = 21! ways.
- There are 22 places of which any 2 places can be filled by those 2 delegates who are never together.
- Two specified delegates can be arranged in \(^{22}P_2\) ways.
- Total number of arrangements if two specified delegates are never together =
\(^{22}P_2 \times 21!\)
\( = \frac{22!}{(22-2)!} \times 21!\)
\( = \frac{22!}{20!} \times 21!\)
\( = 22 \times 21 \times 21!\)
\( = 21 \times 22 \times 21!\)
\( = 21 \times 22!\)
In simple words: If two delegates are always together, treat them as one unit. This unit (23 items in total) can be arranged in 22! ways, multiplied by 2! for internal arrangements. If they are never together, first arrange the remaining 22 delegates in 21! ways, creating 22 gaps, then arrange the two specified delegates in those 22 gaps in \(^{22}P_2\) ways.

🎯 Exam Tip: For "always together" problems, group the specified items and treat them as a single unit, then multiply by the internal permutations of the unit. For "never together" problems, arrange the remaining items first, create gaps, and then place the specified items in those gaps using permutations.

Question 4. Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
Answer:
There are 15 people to sit around a table.
- They can be arranged in (15 - 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e. clockwise and anticlockwise) coincide.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र वृत्ताकार व्यवस्था में बैठे लोगों को दर्शाता है। इसमें दो अलग-अलग व्यवस्थाएँ (घड़ी की दिशा में और घड़ी की विपरीत दिशा में) दिखाई गई हैं, जो दर्शाती हैं कि वृत्ताकार क्रमचय में, यदि पड़ोसियों का क्रम एक समान हो तो ये व्यवस्थाएँ एक ही मानी जाती हैं।
- Number of arrangements possible for not to have same neighbours = \( \frac{14!}{2} \)
In simple words: For 15 people around a table, there are (15-1)! ways. However, if arrangements are considered the same when neighbours are identical (meaning clockwise and anticlockwise arrangements are equivalent), then the total distinct arrangements are (n-1)!/2.

🎯 Exam Tip: When circular permutations specify that arrangements are identical if neighbours are the same (often referred to as 'necklace' type problems or arrangements where objects can be viewed from both sides), divide the standard (n-1)! by 2. Otherwise, use (n-1)!.

Question 5. A committee of 20 members sits around a table. Find the number of arrangements that have the president and the vice president together.
Answer:
A committee of 20 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 18 members of the committee is to be arranged around a table, which can be done in (19 - 1)! = 18! ways.
- The total number of arrangements possible if President and Vice-president sit together = 18! × 2!
In simple words: Treat the President and Vice-president as a single unit. This makes a total of 19 units (18 members + 1 President-Vice President unit) to arrange circularly, which is (19-1)! = 18! ways. Since the President and Vice-president can swap positions within their unit, multiply by 2!.

🎯 Exam Tip: When specific individuals must sit together, combine them into a single 'block' or 'unit'. Calculate the circular permutations for these units, and then multiply by the number of internal arrangements possible within the 'block'.

Question 6. Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated
(i) between the two women.
(ii) between two men.
Answer:
(i) 5 men, 2 women, and a child sit around a table
When a child is seated between two women
- The two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Also, these 3 are to be seated with 5 men, (i.e. a total of 6 units) which can be done in (6 - 1)! = 5! ways.
- The total number of arrangements if the child is seated between two women = 5! × 2!

(ii) Two men out of 5 men can sit on either side of the child in \(^{5}P_2\) ways.
Let us take two men and a child as one unit.
Now these are to be arranged with the remaining 3 men and 2 women
i.e., a total of 6 events (3 + 2 + 1) is to be arranged around a round table which can be done in (6 - 1)! = 5! ways.
- The total number of arrangements, if the child is seated between two men = \(^{5}P_2\) × 5!
In simple words: For (i), treat the two women and child as one unit (W-C-W, 2! internal ways). This unit plus 5 men (total 6 units) can be arranged circularly in (6-1)! ways. For (ii), choose 2 men out of 5 for the child's sides in \(^{5}P_2\) ways, forming one unit. This unit plus the remaining 3 men and 2 women (total 6 units) can be arranged circularly in (6-1)! ways.

🎯 Exam Tip: When a person must sit 'between' two others, form a composite unit. Be careful to distinguish between 'choosing and arranging' (permutation) and 'just arranging' (factorial) for the elements within the unit, based on whether the specific identities of the side-sitters matter or if they are interchangeable.

Question 7. Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Answer:
8 men can be seated around a table in (8 - 1)! = 7! ways.
There are 8 gaps created by 8 men's seats.
- 6 Women can be seated in 8 gaps in \(^{8}P_6\) ways
- Total number of arrangements so that no two women are together = 7! × \(^{8}P_6\)
In simple words: To ensure no two women sit together, first arrange the men circularly in (8-1)! ways. This creates 8 spaces between the men where the 6 women can sit, which can be done in \(^{8}P_6\) ways (choosing 6 out of 8 spaces and arranging the women). The total is the product of these two arrangements.

🎯 Exam Tip: For "no two items together" problems, first arrange the items that *can* be together. This creates spaces. Then, arrange the items that *cannot* be together in those spaces. Use permutations for placing the 'cannot be together' items if their order matters.

Question 8. Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्ताकार मेज को दर्शाता है जिस पर 6 कुर्सियाँ 1 से 6 तक क्रमबद्ध रूप से रखी गई हैं। यह व्यवस्था यह समझने में मदद करती है कि लोग एक वृत्ताकार मेज के चारों ओर कैसे बैठ सकते हैं और विशिष्ट सीटों के संदर्भ में उनकी स्थिति क्या हो सकती है।
Two women sit together and one woman sits separately.
Women sitting separately can be selected in 3 ways.
The other two women occupy two chairs in one way (as it is a circular arrangement).
They can be seated on those two chairs in 2 ways. Suppose two chairs are chairs 1 and 2 shown in the figure.
Then the third woman has only two options viz chairs 4 or 5.
- The third woman can be seated in 2 ways. 3 men are seated in 3! ways
- Required number = 3 × 2 × 2 × 3!
= 12 x 6
= 72
In simple words: To have exactly two women together, first choose which two women sit together (3 ways). Arrange these two women (2 ways). Then, seat the remaining women such that she is not next to the pair. This can be done in 2 distinct ways. The 3 men can then be arranged in the remaining spots in 3! ways. Multiply all these possibilities.

🎯 Exam Tip: When dealing with "exactly N items together" in circular permutations, it often requires careful consideration of both grouping the specified items and placing the remaining items to enforce the 'exactly' condition, preventing more items from joining the group or ensuring the excluded items are sufficiently separated.

Question 9. Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
Answer:
Ten things can be arranged in a circular order of which 4 are alike in \( \frac{9!}{4!} \) ways.
- Required total number of arrangements = \( \frac{9!}{4!} \)
In simple words: For circular arrangements with 'n' objects where 'k' are alike, the number of ways is typically (n-1)!/k! if n and k are coprime, or other formulas apply based on specific properties. Here, for 10 objects with 4 alike, if we fix one distinct object, the remaining 9 can be arranged as if in a line, with 4 alike.

🎯 Exam Tip: Circular permutations with repeated items are complex. If not specified as "necklace" type, the formula (n-1)!/p! (where p is the number of repetitions) is often used for a specific type of setup where one item is fixed. Be careful to check if it's a "bracelet" type (where reversal is considered same), which would halve the result.

Question 10. Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Answer:
Since 2 particular persons can't be sitting side by side.
The other 13 persons can be arranged around the table in (13 - 1)! = 12!
13 people around a table create 13 gaps in which 2 people are to be seated
Number of arrangements of 2 people = \(^{13}P_2\)
- The total number of arrangements in which two specified persons not sitting side by side = 12! × \(^{13}P_2\)
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!
In simple words: To ensure two specific people are never together, first arrange the remaining 13 people circularly in (13-1)! = 12! ways. This creates 13 empty spaces between them. The two specified persons can then be placed in any two of these 13 spaces in \(^{13}P_2\) ways.

🎯 Exam Tip: The "never together" strategy is effective for circular permutations. Arrange the 'other' items first to create 'gaps', then permute the 'restricted' items into these gaps. The number of gaps created by 'n' items in a circle is 'n'.