Permutation and Combination: In how many ways can the letters of the word "BUSINESS" be arranged?

Here are the solutions to the permutation, combination, and probability questions: 26. In how many ways can the letters of the word "BUSINESS" be arranged? Step 1: Count the total number of letters and the frequency of repeated letters. The word "BUSINESS" has 8 letters. The letter 'S' appears 3 times. All other letters (B, U, I, N, E) appear once. Step 2: Use the formula for permutations with repetitions. The number of distinct arrangements of n objects where there are n_1 identical objects of type 1, n_2 identical objects of type 2, ..., is given by (n!)/(n_1! n_2! ...). Here, n=8 and n_S=3. Number of ways = (8!)/(3!) Step 3: Calculate the value. (8!)/(3!) = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1) = 8 × 7 × 6 × 5 × 4 = 6720 The correct option is d) 6720. 27. A committee of 4 members is to be formed from a group of 10 people. In how many ways can this be done? Step 1: Identify this as a combination problem, as the order of members in a committee does not matter. We need to choose 4 members from 10. Step 2: Use the combination formula C(n, k) = nk = (n!)/(k!(n-k)!). Here, n=10 (total people) and k=4 (members to choose). Number of ways = 104 = (10!)/(4!(10-4)!) = (10!)/(4!6!) Step 3: Calculate the value. (10!)/(4!6!) = (10 × 9 × 8 × 7 × 6!)/(4 × 3 × 2 × 1 × 6!) = (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1) = (5040)/(24) = 210 The correct option is a) 210. 28. How many 3-digit numbers can be formed using the digits 1, 2, 3, and 4 without repetition? Step 1: Identify this as a permutation problem, as the order of digits matters for forming different numbers. We have 4 distinct digits (1, 2, 3, 4) and we need to choose and arrange 3 of them. Step 2: Use the permutation formula P(n, k) = (n!)/((n-k)!). Here, n=4 (total digits) and k=3 (digits for the number). Number of ways = P(4, 3) = (4!)/((4-3)!) = (4!)/(1!) Step 3: Calculate the value. (4!)/(1!) = (4 × 3 × 2 × 1)/(1) = 24 The correct option is b) 24. 29. In how many ways can 5 books be arranged on a shelf? Step 1: Identify this as a permutation problem, as the order of books on a shelf matters. We have 5 distinct books to arrange. Step 2: The number of ways to arrange n distinct items is n!. Here, n=5. Number of ways = 5! Step 3: Calculate the value. 5! = 5 × 4 × 3 × 2 × 1 = 120 The correct option is c) 120. 30. A pizza place offers 5 toppings. If a customer can choose any number of toppings, including none, how many different choices are possible? Step 1: For each topping, there are two independent choices: either select it or not select it. Step 2: If there are n items and for each item there are 2 choices, the total number of combinations is 2^n. Here, n=5 (number of toppings). Number of choices = 2^5 Step 3: Calculate the value. 2^5 = 2 × 2 × 2 × 2 × 2 = 32 The correct option is d) 32. 31. In how many ways can 3 different books be arranged on a bookshelf? Step 1: Identify this as a permutation problem, as the order of books on a shelf matters. We have 3 distinct books to arrange. Step 2: The number of ways to arrange n distinct items is n!. Here, n=3. Number of ways = 3! Step 3: Calculate the value. 3! = 3 × 2 × 1 = 6 The correct option is b) 6. 32. How many different 4-letter codes can be formed using letters A, B, C, and D if repetition is allowed? Step 1: For each position in the 4-letter code, there are 4 possible letters (A, B, C, D). Repetition is allowed. Step 2: Since there are 4 positions and 4 choices for each position, the total number of codes is 4 × 4 × 4 × 4. Number of codes = 4^4 Step 3: Calculate the value. 4^4 = 256 The correct option is d) 256. 33. A box contains 6 red balls, 4 blue balls, and 3 green balls. If two balls are drawn randomly, without replacement, what is the probability that both balls are blue? Step 1: Determine the total number of balls and the number of blue balls. Total balls = 6 (red) + 4 (blue) + 3 (green) = 13 balls. Number of blue balls = 4. Step 2: Calculate the probability of drawing the first blue ball. P(1st ball is blue) = Number of blue ballsTotal number of balls = (4)/(13) Step 3: Calculate the probability of drawing the second blue ball, given the first was blue and not replaced. After drawing one blue ball, there are 3 blue balls left and a total of 12 balls remaining. P(2nd ball is blue | 1st ball was blue) = Remaining blue ballsRemaining total balls = (3)/(12) = (1)/(4) Step 4: Multiply the probabilities to find the probability of both events occurring. P(both balls are blue) = P(1st blue) × P(2nd blue | 1st blue) = (4)/(13) × (1)/(4) Step 5: Simplify the result. (4)/(13) × (1)/(4) = (1)/(13) The correct option is e) None of the above. (The calculated probability is (1)/(13), which is not among options a, b, c, d). That's 2 down. 3 left today — send the next one.