Test: Permutations and Combinations- 2 - CA Foundation MCQ

If¹⁸C_r = ¹⁸C_r+2, we can derive the value of ^rC₅ by understanding the properties of binomial coefficients.

• From the equation, we know that C represents combinations. Specifically, C_n^k indicates the number of ways to choose k items from n items.
• Setting ¹⁸C_r equal to ¹⁸C_r+2 implies that the combinations for these two values are equal.
• This equality indicates a specific relationship between r and the total number of items, 18, which helps us find ^rC₅.
• Using the identity of combinations, we can express this as:
• C_n^k = C_n^n-k
• Applying this property, we can simplify the calculations needed to find ^rC₅.

Through this analysis, we conclude that the value of ^rC₅ is 56.

Given the equations:

• nc_r−1 = 56
• nc_r = 28
• nc_r+1 = 8

We need to find the value of r.

To solve this, we can use the property of combinations:

• nc_r = nc_r−1 × r / n−r + 1
• nc_r+1 = nc_r × n−r / r + 1

From the given equations, we can substitute:

• Using nc_r−1 = 56 and nc_r = 28:
• 28 = 56 × r / (n−r+1)
• Thus, we can derive that r = 0.5(n−r+1) = n/2 + 0.5

Next, using nc_r = 28 and nc_r+1 = 8:

• 8 = 28 × (n−r)/(r+1)
• From this, we can derive another relationship.

By solving these equations sequentially, we can find the value of r. Eventually, we find that:

• r = 6

To determine how many ways a person can invite one or more friends from a group of 8, consider the following:

• Each friend can either be invited or not invited.
• This gives us 2 options (invite or not invite) for each friend.
• For 8 friends, the total combinations are calculated as 2⁸.
• This equals 256 combinations, including the case where no friends are invited.
• Since the person must invite at least one friend, we subtract the one combination where no friends are invited.
• The final count of ways to invite one or more friends is 256 - 1 = 255.

Thus, the total number of ways to invite one or more friends is 255.

To determine the number of ways to choose one or more electrical appliances from a set of four options:

• Consider the appliances: T.V., Refrigerator, Washing Machine, and Cooler.
• Each appliance can either be included or excluded in a selection.
• This creates a scenario where each appliance has two options (chosen or not chosen).
• With four appliances, the total number of combinations is calculated as:
• 2ⁿ, where n is the number of appliances.
• For four appliances, this becomes 2⁴ = 16.
• However, this total includes the scenario where no appliances are chosen.
• To find the combinations where at least one appliance is selected, we subtract this one case:
• Thus, the final count is 16 - 1 = 15.

Therefore, there are 15 ways to choose one or more appliances from the given options.

To solve the equationⁿc₁₀ = ⁿc₁₄, we can use the property of combinations.

• First, we know that nc_r gives us the number of ways to choose r objects from n objects.
• The equation can be simplified using the identity: nc_r = nc_n-r.
• This means nc₁₀ = nc₁₄ implies that:
• 10 = n - 14
• So, rearranging gives us: n = 24.

Next, we need to find the value of ²⁵c_n:

• Substituting n = 24 into the equation:
• ²⁵c₂₄ is equal to the number of ways to choose 24 objects from 25.
• This is calculated as:
• ²⁵c₂₄ = 25 (since choosing 24 out of 25 is the same as leaving one object out).

Therefore, the value of ²⁵c_n is 25.

To form a committee of 5 from 7 gentlemen and 4 ladies, ensuring at least one lady is included, we can use the complementary counting method.

First, we calculate the total number of ways to form a committee of 5 from 11 people (7 gents + 4 ladies):

• Total combinations = C(11, 5)

Next, we calculate the number of committees that include no ladies, which means selecting all members from the gents:

• Combinations with no ladies = C(7, 5)

Now, we can find the number of committees that include at least one lady:

• Committees with at least one lady = Total combinations - Committees with no ladies
• This gives us: C(11, 5) - C(7, 5)

Calculating these values:

• C(11, 5) = 462
• C(7, 5) = 21

Thus, the number of committees with at least one lady is:

• 462 - 21 = 441

Therefore, the final answer is 441.

To solve the equation:

• The expression is given as 28C_2r : 24C_{2r – 4} = 225 : 11.
• This implies a ratio that can be simplified.
• Let's express it as a fraction:
• 28C_2r / 24C_{2r – 4} = 225 / 11.
• Cross-multiplying yields:
• 28C_2r × 11 = 225 × 24C_{2r – 4}.
• Next, we need to derive values for C_n:
• C_n is the binomial coefficient, calculated as:
• C_n = n! / (r! × (n - r)!)
• Substituting for C_2r and C_2r-4 gives two equations:
• From the equation, we can equate terms and simplify further, leading us to find the value of r.
• Solving the resulting equation will yield:

The final value of r is 6.

To calculate the number of diagonals in a polygon, you can use the formula:

• The formula for finding the number of diagonals in a polygon with n sides is: Diagonals = n(n - 3) / 2.
• For a decagon, which has 10 sides, substitute n = 10 into the formula:
• Diagonals = 10(10 - 3) / 2
• This simplifies to: Diagonals = 10(7) / 2 = 70 / 2 = 35.

Therefore, a decagon has 35 diagonals.

To determine the number of triangles that can be formed from 12 points in a plane, of which 5 are collinear:

• First, calculate the total combinations of triangles that can be formed by choosing any 3 points from the 12 available.
• This is given by the combination formula: C(n, r) = n! / (r!(n - r)!).
• So, the total combinations for 12 points is: C(12, 3).
• Calculating this gives:
• C(12, 3) = 12! / (3! * 9!) = (12 × 11 × 10) / (3 × 2 × 1) = 220.
• Next, we need to subtract the triangles that cannot be formed due to the collinear points:
• Triangles cannot be formed using all 3 points from the 5 collinear points.
• The number of ways to choose 3 points from these collinear points is:
• C(5, 3) = 5! / (3! * 2!) = (5 × 4) / (2 × 1) = 10.
• Now, subtract the non-triangle forming combinations from the total:
• Total triangles = C(12, 3) - C(5, 3) = 220 - 10 = 210.

Thus, the total number of triangles that can be formed is 210.

The number of straight lines that can be formed by joining 16 points on a plane, with the condition that no two points are collinear, can be calculated using the formula for combinations.

• To find the number of straight lines, we need to select any two points from the 16 available points.
• The formula for combinations is given by: C(n, r) = n! / (r!(n - r)!), where:
• n is the total number of points (16 in this case).
• r is the number of points to choose (2 for a line).
• Applying this to our problem:
• C(16, 2) = 16! / (2!(16 - 2)!)
• This simplifies to: C(16, 2) = 16! / (2! * 14!) = (16 × 15) / (2 × 1) = 120
• Therefore, the total number of straight lines that can be formed is 120.

To determine the number of ways a voter can vote:

• A voter can select candidates from a total of 5 candidates.
• They can vote for any number of candidates, up to a maximum of 3.

To find the total number of voting combinations, we can break it down as follows:

• Voting for 0 candidates: 1 way (not voting).
• Voting for 1 candidate: There are 5 ways (one for each candidate).
• Voting for 2 candidates: The number of ways to choose 2 out of 5 can be calculated using combinations:
• Combinations of 5 choose 2 = 10.
• Voting for 3 candidates: The number of ways to choose 3 out of 5 is:
• Combinations of 5 choose 3 = 10.

Now, we can sum up all the possible ways a voter can choose:

• 0 candidates: 1 way
• 1 candidate: 5 ways
• 2 candidates: 10 ways
• 3 candidates: 10 ways

Total ways:1 + 5 + 10 + 10 = 26

Therefore, the number of ways a voter can choose to vote is 26.

To determine the number of guests at the party based on the total handshakes:

In a party, every guest shakes hands with every other guest. The total number of handshakes can be calculated using the formula:

• Handshakes = n(n - 1) / 2, where n is the number of guests.

Given that the total number of handshakes is 66, we can set up the equation:

• n(n - 1) / 2 = 66

To eliminate the fraction, multiply both sides by 2:

• n(n - 1) = 132

Now, we need to find two consecutive integers whose product is 132. This can be done by trial and error:

• If n = 12, then 12 x 11 = 132.
• If n = 11, then 11 x 10 = 110 (too low).
• If n = 13, then 13 x 12 = 156 (too high).

Thus, the only solution that fits is n = 12. Therefore, the number of guests at the party is:

• 12

The problem involves determining the number of parallelograms that can be formed from two sets of parallel lines. In this case, we have:

• Four parallel lines intersecting
• Three parallel lines

To find the total number of parallelograms, we can follow these steps:

• Select two lines from the first set of four lines.
• Select two lines from the second set of three lines.

The total number of ways to choose two lines from a set of n lines is given by the combination formula:

Combination (n, r) = n! / [r!(n - r)!]

Applying this to our sets:

• From the first set (4 lines):
• Combination (4, 2) = 6
• From the second set (3 lines):
• Combination (3, 2) = 3

Now, to find the total number of parallelograms, we multiply the two results:

Total Parallelograms = 6 (from first set) * 3 (from second set) = 18

Thus, the number of parallelograms that can be formed is 18.

The number of ways to divide 12 students into three equal groups is calculated using combinatorial mathematics. Here’s a simplified breakdown of the solution:

• To form three groups of four from 12 students, we use the formula for combinations.
• First, select 4 students from the 12. The number of ways to do this is given by:
• C(12, 4) = 495
• Next, select another 4 students from the remaining 8:
• C(8, 4) = 70
• The last 4 students automatically form the final group.
• Since the order of the groups does not matter, we must divide by the number of arrangements of the three groups, which is:
• 3! = 6

Putting this together, the total number of ways to divide the students is calculated as follows:

• Total Ways = (C(12, 4) × C(8, 4)) / 3!
• Total Ways = (495 × 70) / 6
• Total Ways = 34650 / 6 = 5775

Thus, the final result indicates that there are 5775 ways to divide 12 students into three equal groups.

To determine the number of ways to divide 15 mangoes equally among 3 students, we can follow these steps:

• The total number of mangoes is 15.
• Since there are 3 students, we need to divide the mangoes by the number of students.
• Each student would receive:
• 15 mangoes ÷ 3 students = 5 mangoes per student.
• Now, we need to consider the different ways to distribute the mangoes.
• Since each student receives the same number of mangoes, the arrangement is straightforward:
• There is only 1 way to equally distribute 15 mangoes among 3 students, as each gets 5.

Conclusion: The number of ways to equally divide 15 mangoes among 3 students is 1.

To calculate the number of chords formed by joining pairs of points on a circle, we can use combinatorial mathematics.

• The formula for finding the number of ways to choose 2 points from n points is given by:
• C(n, 2) = n! / (2!(n - 2)!)
• In this case, n is 8 since there are 8 points on the circumference.
• Substituting the values into the formula:
• C(8, 2) = 8! / (2!(8 - 2)!) = (8 × 7) / (2 × 1) = 28
• Thus, the total number of chords that can be formed is 28.

Therefore, the answer is 28.

To form a committee of 3 ladies and 4 gents from a pool of 8 ladies and 7 gents, we must account for the condition that Mrs. X will not serve if Mr. Y is also on the committee.

We can approach this problem in two parts:

• Case 1: Committees that do not include Mr. Y.
• Case 2: Committees that include Mr. Y but not Mrs. X.

Case 1: If Mr. Y is excluded, we can choose from 6 gents. The calculations are as follows:

• Choose 3 ladies from 8: C(8, 3) = 56
• Choose 4 gents from 6: C(6, 4) = 15
• Total for this case: 56 x 15 = 840

Case 2: If Mr. Y is included, Mrs. X cannot be. We choose from 7 ladies and 6 gents:

• Choose 3 ladies from 7: C(7, 3) = 35
• Choose 3 gents from 6: C(6, 3) = 20
• Total for this case: 35 x 20 = 700

Now, add the totals from both cases:

• Case 1 total: 840
• Case 2 total: 700
• Overall total = 840 + 700 = 1540

Thus, the total number of valid committees is 1540.

To solve the equation:

• The equation can be rewritten as: C(500, 92) = C(499, 92) + C(n, 91).
• This represents a combinatorial identity where the left side counts the ways to choose 92 items from 500.
• The right side counts two scenarios: choosing 92 from 499 and choosing 91 from a set represented by n.
• According to the identity, n must be 500 because:
• When we choose 91 from n, we need to account for the one item left out from 500.
• Thus, n must equal 500 to maintain balance in the equation.
• Therefore, x equals 500, confirming that the solution is accurate.

The Supreme Court's decision was a 6 to 3 ruling that upheld the decision of a lower court. To consider how many ways a majority decision could potentially reverse this ruling, it is important to understand the voting structure.

• The Supreme Court consists of nine justices.
• A majority decision requires at least five votes to reverse the lower court's ruling.
• When justices vote, each can either support or oppose the reversal.

To calculate the number of combinations for a majority decision:

• We look for all possible voting patterns where at least five justices are in favour of reversing the decision.
• This involves considering scenarios like:
• 5 justices voting for reversal
• 6 justices voting for reversal
• 7 justices voting for reversal
• 8 justices voting for reversal
• 9 justices voting for reversal

The total combinations can be calculated using the binomial coefficient, which gives us the possible ways justices can vote in these scenarios. This results in:

• 5 votes for reversal: C(9,5)
• 6 votes for reversal: C(9,6)
• 7 votes for reversal: C(9,7)
• 8 votes for reversal: C(9,8)
• 9 votes for reversal: C(9,9)

Summing these combinations will yield the total number of ways the Court can reach a majority decision to reverse the lower court's ruling. The final calculation leads to 256 unique ways.

To determine the number of trials needed to light the room, we consider the following points:

• There are a total of five bulbs, out of which three are defective.
• Only two bulbs can be tested at a time in the bulb points.
• The goal is to find out how many different combinations of bulbs can be tried.

To calculate the number of trials:

• We can select any two bulbs from the five available.
• The number of combinations can be calculated using the formula for combinations: C(n, r) = n! / [r! * (n - r)!], where n is the total number of bulbs and r is the number of bulbs to choose.
• In this case, we need to calculate C(5, 2).
• This gives us: C(5, 2) = 5! / [2! * (5 - 2)!] = 10.

However, since the question specifically asks about the number of trials in which the room will be lit:

• We can only light the room when both bulbs are non-defective.
• From the five bulbs, there are two functional bulbs.
• The combinations of the two working bulbs can only be tested once.

Thus, the total number of trials during which the room will be lit is 1, as we can only have one combination that successfully lights the room.

CALCUTTA has 8 letters, with the letter 'T' appearing twice. The formula for calculating the total arrangements of its letters is:

• Total arrangements = n! / (p1! * p2!), where:
• n = total letters, p1 = count of letter 'T'.

Thus, the total arrangements for CALCUTTA is:

• Total = 8! / 2! = 40320 / 2 = 20160 arrangements.

AMERICA has 7 letters, with the letter 'A' appearing twice. The formula for its arrangements is similar:

• Total arrangements = n! / (p1!), where:
• n = total letters, p1 = count of letter 'A'.

Therefore, the total arrangements for AMERICA is:

• Total = 7! / 2! = 5040 / 2 = 2520 arrangements.

To find the ratio of the arrangements of CALCUTTA to AMERICA:

• Ratio = 20160 : 2520.
• Simplifying this gives a final ratio of 8 : 1.

Since the answer choices do not include 8 : 1, the correct response is none of these.

To find the ways of selecting 4 letters from the word EXAMINATION:

• The word EXAMINATION consists of 11 letters.
• It contains the letters: E, X, A, M, I, N, A, T, I, O, N.
• Notably, the letters A and I are repeated.

The total number of distinct letters is:

• Unique letters: E, X, A, M, I, N, T, O (total of 8 unique letters).

To calculate the number of ways to select 4 letters, consider two scenarios:

• All letters different: Choose any 4 from the 8 unique letters.
• Including repetitions: Consider combinations that include repeated letters A or I.

Using combinations, the calculations can be broken down as follows:

• Choosing 4 different letters from 8: C(8, 4).
• Calculating C(8, 4) gives 70 ways.
• For cases with one letter repeated, further combinations must be analysed.

Considering the repetition of letters A and I:

• If A is chosen twice, select 2 more from the remaining 7 letters: C(7, 2).
• If I is chosen twice, similarly select 2 from the remaining letters.

By adding all valid combinations together, you can derive the total number of ways.

The number of different words that can be formed using 12 consonants and 5 vowels by selecting 4 consonants and 3 vowels is calculated as follows:

• First, calculate the combinations of consonants:
• 12 choose 4 is represented as ¹²c₄.
• Next, calculate the combinations of vowels:
• 5 choose 3 is represented as ⁵c₃.
• Multiply these two results together:
• The total number of combinations is 12c4 × 5c3.

This formula provides the total number of unique combinations of words that can be formed with the specified consonants and vowels.

To determine the number of seating arrangements for eight guests at a long rectangular table, we consider the seating preferences of certain guests:

• Two guests want to sit on one side of the table.
• Three guests prefer to sit on the opposite side.

The remaining guests can be seated based on these preferences.

Here’s how we calculate the total seating arrangements:

• First, we choose positions for the two guests on one side. There are 4 spots available on that side.
• The number of ways to arrange these two guests is given by the combinations of 4 taken 2 at a time: C(4, 2).
• Next, we arrange the two chosen guests in their seats, which can be done in 2! ways.

For the other side of the table:

• We select 3 out of the 4 available seats for the three guests who wish to sit there.
• The number of ways to choose these seats is also C(4, 3).
• We can arrange these three guests in their seats in 3! ways.

Finally, the last two guests can occupy the remaining seats:

• They can be seated in 2! different ways.

Now, combining all these calculations gives us:

• Total arrangements = C(4, 2) × 2! × C(4, 3) × 3! × 2!
• Calculating each part:
• C(4, 2) = 6, C(4, 3) = 4, therefore, the total becomes: 6 × 2 × 4 × 6 × 2 = 1728.

Thus, the total number of ways the guests can be seated is 1728.