1. Explain the principle of multiplication in permutations and combinations and provide an example.
Answer: The principle of multiplication states that if there are m ways to do one thing and n ways to do another thing, then there are m x n ways to do both things. In permutations, this principle is applied when we have multiple choices for each position in the arrangement. For example, if we have 3 choices for the first position and 4 choices for the second position, then there are 3 x 4 = 12 possible arrangements.
2. Prove the formula for permutations of n objects taken r at a time using the factorial notation.
Answer: The formula for permutations of n objects taken r at a time is given by P(n, r) = n! / (n – r)!. To prove this formula, we start with the definition of factorial: n! = n x (n-1) x (n-2) x … x 2 x 1. Now, let’s consider the number of ways we can arrange r objects out of n. We have n choices for the first object, (n-1) choices for the second object, and so on, until we have (n-r+1) choices for the rth object. Therefore, the total number of arrangements is n x (n-1) x (n-2) x … x (n-r+1), which is equal to n! / (n-r)!. Hence, the formula is proved.
3. Discuss the concept of circular permutations and provide an example.
Answer: Circular permutations occur when objects are arranged in a circle, and the order matters. In circular permutations, the relative order of the objects is important, but the starting point is not. For example, consider arranging 4 people in a circular table. The first person can be chosen in 4 ways. After fixing the first person, the second person can be chosen in 3 ways (as one seat is already occupied). Similarly, the third person can be chosen in 2 ways, and the last person is automatically fixed. Therefore, the total number of circular permutations is 4 x 3 x 2 = 24.
4. Explain the concept of combinations and provide an example.
Answer: Combinations are selections of objects where the order does not matter. In combinations, the selection is made without considering the arrangement or order of the objects. For example, consider selecting 3 students from a group of 10 students to form a committee. The order in which the students are selected does not matter. The number of combinations can be calculated using the formula C(n, r) = n! / (r! x (n-r)!), where n is the total number of objects and r is the number of objects to be selected. In this case, the number of combinations is C(10, 3) = 10! / (3! x (10-3)!) = 120.
5. Discuss the concept of the binomial theorem and provide an example.
Answer: The binomial theorem is a formula that allows us to expand the powers of a binomial. It states that for any positive integer n, the expansion of (a + b)^n can be expressed as the sum of terms of the form C(n, r) x a^(n-r) x b^r, where C(n, r) represents the number of combinations of n objects taken r at a time. For example, consider expanding (x + y)^4. Using the binomial theorem, we have (x + y)^4 = C(4, 0) x x^4 x y^0 + C(4, 1) x x^3 x y^1 + C(4, 2) x x^2 x y^2 + C(4, 3) x x^1 x y^3 + C(4, 4) x x^0 x y^4. Simplifying this expression gives x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4.
6. Prove the formula for the sum of combinations using the principle of addition.
Answer: The formula for the sum of combinations is given by C(n, 0) + C(n, 1) + C(n, 2) + … + C(n, n) = 2^n. To prove this formula, we can use the principle of addition. Consider a set of n objects, and we want to count the number of ways to select any number of objects from the set. This can be done by either including an object in the selection or excluding it. Therefore, the total number of selections is equal to the sum of the number of selections when the object is included (C(n, r)) and when the object is excluded (C(n, r-1)). Simplifying this expression gives C(n, 0) + C(n, 1) + C(n, 2) + … + C(n, n) = 2^n.
7. Discuss the concept of repeated permutations and provide an example.
Answer: Repeated permutations occur when there are objects with repeated elements, and the order matters. In repeated permutations, the total number of arrangements is calculated by multiplying the number of choices for each position. For example, consider arranging the letters of the word “MISSISSIPPI.” Here, we have 11 letters, but there are repeated letters: 4 “I”s, 4 “S”s, and 2 “P”s. The total number of arrangements can be calculated as 11! / (4! x 4! x 2!) = 34,650.
8. Explain the concept of derangements and provide an example.
Answer: Derangements refer to arrangements in which none of the objects occupy their original positions. In other words, it is a permutation with no fixed points. For example, consider arranging the letters A, B, and C such that none of the letters are in their original positions. The possible derangements are BCA and CAB. The formula for calculating the number of derangements of n objects is given by D(n) = n! x (1 – 1/1! + 1/2! – 1/3! + … + (-1)^n/n!).
9. Discuss the concept of multinomial coefficients and provide an example.
Answer: Multinomial coefficients are used when objects are divided into several groups, and the order within each group matters. The formula for multinomial coefficients is given by C(n; n1, n2, …, nk) = n! / (n1! x n2! x … x nk!), where n is the total number of objects and n1, n2, …, nk represent the number of objects in each group. For example, consider dividing 6 objects into groups of sizes 2, 2, and 2. The number of arrangements can be calculated as C(6; 2, 2, 2) = 6! / (2! x 2! x 2!) = 90.
10. Prove the formula for combinations with repetitions using the concept of stars and bars.
Answer: The formula for combinations with repetitions is given by C(n + r – 1, r) = (n + r – 1)! / (r! x (n – 1)!), where n is the total number of objects and r is the number of objects to be selected. To prove this formula, we can use the concept of stars and bars. Imagine representing the objects as stars and using bars to divide them into groups. The number of ways to arrange the stars and bars is equal to the number of combinations with repetitions. By counting the number of arrangements, we can derive the formula mentioned above.