Samacheer Kalvi Class 11 Business Maths Solutions Chapter 2 Algebra Exercise 2.4

Question 1. If \( n P_r = 1680 \) and \( n C_r = 70 \), find n and r.
Answer: Given that \( n P_r = 1680 \) and \( n C_r = 70 \). We know the relationship between permutation and combination is \( n C_r = \frac{n P_r}{r!} \).
Substitute the given values into the formula:
\( 70 = \frac{1680}{r!} \)
Now, solve for \( r! \):
\( r! = \frac{1680}{70} \)
\( r! = 24 \)
We know that \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). Therefore, \( r = 4 \).
Next, use the value of \( r \) in the permutation formula \( n P_r = 1680 \):
\( n P_4 = 1680 \)
We need to find four consecutive descending numbers starting from \( n \) whose product is 1680. We can test values for \( n \).
Let's try \( n = 8 \):
\( 8 \times 7 \times 6 \times 5 = 56 \times 30 = 1680 \). This matches.
So, \( n = 8 \).
Thus, the values are \( n = 8 \) and \( r = 4 \). Permutations show the arrangement, and combinations show selection.
In simple words: We used a formula that links combinations and permutations to find 'r'. Then, we used the permutation formula with 'r' to find 'n'. The values found are n equals 8 and r equals 4.

🎯 Exam Tip: Always remember the fundamental relation \( n C_r = \frac{n P_r}{r!} \) as it often simplifies solving problems involving both permutations and combinations. You can test small integer values to find 'n' for permutation equations.

Question 2. Verify that \( 8 C_4 + 8 C_3 = 9 C_4 \).
Answer: We need to verify the given equation: \( 8 C_4 + 8 C_3 = 9 C_4 \).
This equation is an example of Pascal's Identity, which states \( n C_r + n C_{r-1} = (n+1) C_r \).
In this case, \( n = 8 \) and \( r = 4 \). So, the identity directly applies:
\( 8 C_4 + 8 C_{4-1} = (8+1) C_4 \)
\( 8 C_4 + 8 C_3 = 9 C_4 \). This verifies the statement directly.

Alternatively, we can calculate each term:
Calculate the Left Hand Side (LHS):
\( 8 C_4 = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \)
\( 8 C_3 = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \)
LHS = \( 70 + 56 = 126 \).

Calculate the Right Hand Side (RHS):
\( 9 C_4 = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126 \).
Since LHS = RHS (126 = 126), the equation is verified. Pascal's Identity helps us understand how combination values build upon each other in a systematic way.
In simple words: We can verify this by using Pascal's rule, which shows that adding two specific combinations gives the next one. Also, calculating both sides separately shows they both equal 126, proving the statement is true.

🎯 Exam Tip: Recognizing Pascal's Identity can save time in verification problems. If you forget the identity, calculating each combination individually is a reliable alternative.

Question 3. How many chords can be drawn through 21 points on a circle?
Answer: To draw a single chord on a circle, we need to choose exactly two distinct points on the circle. The order in which we choose the points does not matter, so this is a combination problem.
We have 21 points available on the circle, and we need to choose 2 points at a time.
Using the combination formula \( n C_r = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of points and \( r \) is the number of points to choose for each chord.
Here, \( n = 21 \) and \( r = 2 \).
Number of chords = \( 21 C_2 \)
\( 21 C_2 = \frac{21 \times (21-1)}{2 \times 1} \)
\( = \frac{21 \times 20}{2} \)
\( = 21 \times 10 \)
\( = 210 \).
Therefore, 210 chords can be drawn through 21 points on a circle. Every pair of points forms a unique chord.
In simple words: To make a chord, you need two points. Since the order doesn't matter, we use combinations to pick 2 points out of 21. This gives us 210 possible chords.

🎯 Exam Tip: Always identify if a problem is about permutations (order matters) or combinations (order does not matter). For drawing lines or forming groups, it's usually combinations.

Question 4. How many triangles can be formed by joining the vertices of a hexagon?
Answer: A hexagon is a polygon with 6 vertices (corners). To form a triangle, we need to choose any three distinct vertices from these available vertices. The order in which the vertices are chosen does not change the triangle formed, so this is a combination problem.
We have 6 vertices, and we need to choose 3 vertices at a time.
Using the combination formula \( n C_r = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of vertices and \( r \) is the number of vertices needed for a triangle.
Here, \( n = 6 \) and \( r = 3 \).
Number of triangles = \( 6 C_3 \)
\( 6 C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \)
\( = \frac{120}{6} \)
\( = 20 \).
Therefore, 20 distinct triangles can be formed by joining the vertices of a hexagon. This concept can be applied to any polygon to find the number of triangles that can be formed from its vertices.
In simple words: A hexagon has 6 corners. To make a triangle, you pick 3 of these corners. Since the order doesn't matter, there are 20 different triangles you can make.

🎯 Exam Tip: For problems involving forming geometric shapes (like triangles or polygons) by connecting points, combinations are typically used because the order of selecting points does not change the resulting shape.

Question 5. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Answer: This problem involves two main steps: first, selecting the letters (combination), and second, arranging those selected letters to form words (permutation).

**Step 1: Selection of letters**
We need to choose 3 consonants from 7 available consonants. The number of ways to do this is given by \( 7 C_3 \).
\( 7 C_3 = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \) ways.

Next, we need to choose 2 vowels from 4 available vowels. The number of ways to do this is given by \( 4 C_2 \).
\( 4 C_2 = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \) ways.

The total number of ways to select 3 consonants and 2 vowels is the product of the individual selections:
Total selections = \( 7 C_3 \times 4 C_2 = 35 \times 6 = 210 \) ways.

**Step 2: Arrangement of selected letters**
Once we have selected a group of 5 letters (3 consonants + 2 vowels), we need to arrange them to form a word. The number of ways to arrange 5 distinct letters is given by \( 5! \).
\( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \) ways.

**Step 3: Total number of words**
To find the total number of words that can be formed, we multiply the total number of selections by the total number of arrangements for each selection:
Total words = (Total selections) \( \times \) (Arrangement of 5 letters)
Total words = \( 210 \times 120 = 25200 \).
Therefore, 25,200 words can be formed. Many real-world problems combine selection and arrangement steps.
In simple words: First, we pick 3 consonants out of 7, and 2 vowels out of 4. This gives us 210 sets of letters. Then, for each set of 5 letters, we find all the ways to put them in order, which is 120 ways. Multiplying these together gives the total number of words, which is 25,200.

🎯 Exam Tip: When forming "words" or "arrangements" from a selection, remember that it's often a two-step process: first choose the items (combination), then arrange them (permutation).

Question 6. If four dice are rolled, find the number of possible outcomes in which atleast one die shows 2.
Answer: To find the number of outcomes where at least one die shows a 2, it's easier to use the complementary method. This means we calculate the total possible outcomes and subtract the outcomes where *no* die shows a 2.

**Step 1: Calculate total possible outcomes**
When a single die is rolled, there are 6 possible outcomes (1, 2, 3, 4, 5, 6).
When four dice are rolled, the total number of possible outcomes is \( 6 \times 6 \times 6 \times 6 = 6^4 = 1296 \).

**Step 2: Calculate outcomes where no die shows 2**
If a die does *not* show a 2, then there are 5 possible outcomes for that die (1, 3, 4, 5, 6).
If four dice are rolled and none of them show a 2, the number of outcomes is \( 5 \times 5 \times 5 \times 5 = 5^4 = 625 \).

**Step 3: Calculate outcomes where at least one die shows 2**
Number of outcomes (at least one 2) = (Total outcomes) - (Outcomes where no die shows 2)
\( = 1296 - 625 \)
\( = 671 \).
Therefore, there are 671 possible outcomes in which at least one die shows 2. This approach is very useful for "at least" type probability problems.
In simple words: First, we find all possible results from rolling four dice. Then, we find all results where the number 2 does not appear on any die. Subtracting the second number from the first gives us the results where at least one die shows the number 2.

🎯 Exam Tip: For problems involving "at least one," it's often simpler to use the complementary approach: calculate the total possibilities and subtract the possibilities where the event *doesn't* occur at all.

Question 7. There are 18 guests at a dinner party. They have to sit 9 guests on either side of a long table, three particular persons decide to sit on one side and two others on the other side. In how many ways can the guests be seated?
Answer: Total number of guests = 18.
The table has two sides (let's call them Side A and Side B), with 9 seats on each side.

First, consider the particular guests:
* 3 particular persons insist on sitting on Side A.
* 2 other particular persons insist on sitting on Side B.
This means 5 particular guests have already chosen their side.

Number of remaining guests = \( 18 - 5 = 13 \).

Now, let's determine the remaining seats on each side:
* Side A needs \( 9 - 3 = 6 \) more guests.
* Side B needs \( 9 - 2 = 7 \) more guests.

We need to distribute the 13 remaining guests into these remaining seats.
We can choose 6 guests for Side A from the 13 remaining guests. This can be done in \( 13 C_6 \) ways.
\( 13 C_6 = \frac{13!}{6!(13-6)!} = \frac{13!}{6!7!} \).
Once 6 guests are chosen for Side A, the remaining \( 13 - 6 = 7 \) guests will automatically sit on Side B (which requires 7 guests), so there is only \( 7 C_7 = 1 \) way to choose them.
So, the number of ways to select guests for the sides is \( 13 C_6 \).

After selecting the guests for each side, we need to arrange them in their seats:
* On Side A, there are 9 guests (3 particular + 6 chosen). These 9 guests can be arranged in \( 9! \) ways.
* On Side B, there are 9 guests (2 particular + 7 remaining). These 9 guests can also be arranged in \( 9! \) ways.

Therefore, the total number of ways the guests can be seated is:
Total ways = (Ways to choose guests for sides) \( \times \) (Arrangements on Side A) \( \times \) (Arrangements on Side B)
Total ways = \( 13 C_6 \times 9! \times 9! \).
This problem involves both combinations (for selecting which guests go to which side) and permutations (for arranging them in seats).
In simple words: First, we place the special guests. Then, from the remaining 13 guests, we choose 6 for one side of the table. The other 7 automatically go to the other side. Finally, we multiply this by all the ways the 9 guests on each side can sit in their specific seats.

🎯 Exam Tip: Break down complex seating arrangement problems into selection (combinations) and arrangement (permutations) steps. Handle fixed positions or guest preferences first, then consider the remaining choices.

Question 8. If a polygon has 44 diagonals, find the number of its sides.
Answer: Let \( n \) be the number of sides of the polygon. A polygon with \( n \) sides also has \( n \) vertices.
To form any line segment by joining two vertices of the polygon, we need to choose 2 vertices out of \( n \). The number of ways to do this is given by the combination formula \( n C_2 \).
These line segments can either be sides of the polygon or its diagonals.
The number of sides of the polygon is \( n \).
So, the number of diagonals is the total number of line segments minus the number of sides:
Number of diagonals = \( n C_2 - n \)
We are given that the number of diagonals is 44.
So, \( n C_2 - n = 44 \)
Substitute the formula for \( n C_2 \): \( \frac{n(n-1)}{2} - n = 44 \)
Multiply the term \( n \) by \( \frac{2}{2} \) to get a common denominator:
\( \frac{n(n-1) - 2n}{2} = 44 \)
Simplify the numerator:
\( \frac{n^2 - n - 2n}{2} = 44 \)
\( \frac{n^2 - 3n}{2} = 44 \)
Multiply both sides by 2:
\( n^2 - 3n = 88 \)
Rearrange the equation into a standard quadratic form:
\( n^2 - 3n - 88 = 0 \)
Now, we solve this quadratic equation for \( n \). We can factorize the quadratic expression. We look for two numbers that multiply to -88 and add up to -3. These numbers are -11 and 8.
\( (n - 11)(n + 8) = 0 \)
This gives two possible values for \( n \):
\( n - 11 = 0 \implies n = 11 \)
\( n + 8 = 0 \implies n = -8 \)
Since the number of sides of a polygon cannot be negative, we discard \( n = -8 \).
Therefore, the number of sides of the polygon is 11. This polygon is called an hendecagon or undecagon.
In simple words: We used a formula that connects the number of sides of a polygon to its number of diagonals. We set this formula equal to 44 (the given number of diagonals) and solved the resulting equation. Since a polygon can't have negative sides, the answer is 11.

🎯 Exam Tip: Remember the formula for the number of diagonals in an \( n \)-sided polygon: \( \frac{n(n-3)}{2} \). Using this formula directly can save a step in solving the problem.

Question 9. In how many ways can a cricket team of 11 players be chosen out of a batch of 15 players?
(i) no restriction on the selection.
(ii) A particular player is always chosen.
(iii) A particular player is never chosen.
Answer: We have a total of 15 players, and we need to choose a cricket team of 11 players. This is a combination problem because the order of choosing players does not matter.

(i) **No restriction on the selection:**
We need to choose 11 players from 15 without any specific conditions.
Number of ways = \( 15 C_{11} \)
Using the property \( n C_r = n C_{n-r} \), we can simplify the calculation:
\( 15 C_{11} = 15 C_{(15-11)} = 15 C_4 \)
\( 15 C_4 = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} \)
\( = 15 \times 7 \times 13 \) (after cancelling 4, 3, 2 with 12, 14)
\( = 1365 \) ways.

(ii) **A particular player is always chosen:**
If one specific player must always be in the team, then we effectively have 10 remaining spots to fill in the team (11 - 1 = 10).
The pool of players to choose from is now 14 (15 total players - 1 particular player already chosen = 14).
So, we need to choose 10 players from these 14 remaining players.
Number of ways = \( 14 C_{10} \)
Again, using \( n C_r = n C_{n-r} \):
\( 14 C_{10} = 14 C_{(14-10)} = 14 C_4 \)
\( 14 C_4 = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} \)
\( = 7 \times 13 \times 11 \) (after cancelling 4, 3, 2 with 12, 14)
\( = 1001 \) ways.

(iii) **A particular player is never chosen:**
If one specific player must never be in the team, then we exclude this player from the initial pool of 15 players. The team still needs 11 players.
The pool of players to choose from is now 14 (15 total players - 1 particular player excluded = 14).
So, we need to choose 11 players from these 14 remaining players.
Number of ways = \( 14 C_{11} \)
Using \( n C_r = n C_{n-r} \):
\( 14 C_{11} = 14 C_{(14-11)} = 14 C_3 \)
\( 14 C_3 = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} \)
\( = 14 \times 13 \times 2 \) (after cancelling 3, 2, 1 with 12)
\( = 364 \) ways.
In simple words: For the first part, we choose 11 players from 15 without any rules. For the second part, one player is fixed, so we choose the remaining 10 from 14 players. For the third part, one player is not allowed, so we choose all 11 players from the remaining 14 players.

🎯 Exam Tip: Conditional combination problems require adjusting both the total pool of items (n) and the number of items to choose (r) before applying the combination formula.

Question 10. A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways this can be done when
(i) atleast two ladies are included.
(ii) atmost two ladies are included.
Answer: We have 6 gents and 4 ladies. A committee of 5 members is to be formed.

(i) **At least two ladies are included:**
"At least two ladies" means the committee must have 2, 3, or 4 ladies. We cannot have more than 4 ladies because there are only 4 available.
We will consider each case separately and sum the results.

Total ways for (i) = \( 120 + 60 + 6 = 186 \) ways.

(ii) **At most two ladies are included:**
"At most two ladies" means the committee can have 0, 1, or 2 ladies. We will consider each case separately and sum the results.

Total ways for (ii) = \( 6 + 60 + 120 = 186 \) ways.
In simple words: For the first part, we add up the ways to form a committee with 2, 3, or 4 ladies. For the second part, we add up the ways to form a committee with 0, 1, or 2 ladies. Interestingly, both conditions result in 186 possible ways to form the committee.

🎯 Exam Tip: Always break down "at least" or "at most" conditions into individual cases that cover all possibilities. Ensure that the total number of members chosen (ladies + gents) always equals the required committee size.