The word BALLOON has 7 letters with L repeating twice and O repeating twice. Total arrangements = 7! / (2! * 2!) = 1260. Arrangements where Ls are together (treating LL as one unit) = 6! / 2! = 360. Subtracting the cases where Ls are together from the total gives 1260 - 360 = 900 valid arrangements.
532
How many distinct necklaces can be created using 6 beads of different colors?
For arranging n distinct items in a circle, the number of arrangements is (n-1)!. For a necklace, since flipping the necklace does not create a new arrangement, we divide by 2. Thus, (6-1)! / 2 = 120 / 2 = 60 distinct necklaces.
533
In how many distinct ways can 10 individuals arrange themselves in a line at a cinema ticket window?
The number of ways to arrange n distinct objects in a line is given by n factorial (n!). For 10 people, the calculation is 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, which equals 3,628,800.
534
In how many distinct ways can four unique books, one each in Chemistry, Physics, Biology, and Mathematics, be arranged on a bookshelf?
To arrange n distinct objects in a linear sequence, we use the factorial function. For four distinct books, the number of possible arrangements is calculated as 4! (4 factorial), which is 4 × 3 × 2 × 1 = 24. Therefore, there are 24 unique ways to order these books on a shelf.
535
In how many distinct ways can 3 unique positions be filled if there are 6 candidates available for selection?
Since the posts are distinct, the order of selection matters. We use the permutation formula P(n, r) = n! / (n-r)!. Here, n=6 and r=3, so P(6, 3) = 6 × 5 × 4 = 120. This calculation accounts for all possible arrangements of candidates into the three specific posts.
536
In how many distinct ways can a cricket team of 11 players be selected from a pool of 16 players, given that one specific player must always be excluded?
Since one specific player is excluded, we must select 11 players from the remaining 15 players. The number of ways to choose 11 items from 15 is calculated using the combination formula C(n, r) = n! / (r!(n-r)!). Here, C(15, 11) = 15! / (11! * 4!) = (15 * 14 * 13 * 12) / (4 * 3 * 2 * 1) = 1365.
537
In a race involving ten participants, calculate the total number of distinct ways the first, second, and third place prizes can be awarded.
Since the order of the first three positions matters, we use permutations. The number of ways to arrange 3 winners out of 10 participants is calculated as P(10, 3) = 10 × 9 × 8 = 720. This accounts for the specific assignment of gold, silver, and bronze medals to the top three finishers.
538
Determine the missing value in the following numerical analogy: 42 : 56 :: 72 : ?
The pattern follows the product of consecutive integers: 42 is 6 multiplied by 7, and 56 is 7 multiplied by 8. Similarly, 72 is 8 multiplied by 9. Following this logical progression, the next term should be 9 multiplied by 10, which equals 90.
539
Determine the next number in the sequence: 2, 3, 8, 27, 112, ...
Analyze the differences between consecutive terms: 10-3=7, 29-10=19, 66-29=37, 127-66=61. The second differences are 19-7=12, 37-19=18, 61-37=24. These second differences increase by 6 each time. The next second difference is 24+6=30. Adding 30 to the last first difference (61+30=91) gives the next increment. Finally, 127+91=218. Alternatively, the sequence follows the pattern n^3 + 2.
