The newly elected officers of a subdivision are planning to produce a house number made up of 2 letters and 3 digits. The letters to be used are from A to E only and the digits are from 0 to 5 with no repetition. These will be sold to the members at P100 each to recover the expenses and to raise some funds. If all house numbers are taken, how much money will they receive?
There are 5 choices for the first letter and 4 choices for the second letter (since no repetition is allowed). For the first digit, there are 6 choices, for the second digit there are 5 choices, and for the third digit there are 4 choices (again, no repetition is allowed). Therefore, the total number of possible house numbers is:
5 x 4 x 6 x 5 x 4 = 2,400
If they sell all house numbers at P100 each, they will receive:
2,400 x P100 = P240,000
Therefore, they will receive P240,000 if all house numbers are taken.
To determine the total number of possible house numbers, we need to calculate the permutations of the available letters and digits.
For the letters, there are 5 options (A, B, C, D, E) and we need to choose 2 letters without repetition. So, we use the formula for permutations:
P(n, r) = n! / (n - r)!
In this case, n = 5 (number of available options for letters) and r = 2 (number of letters to choose):
P(5, 2) = 5! / (5 - 2)! = 5! / 3! = (5 x 4 x 3!) / 3! = 5 x 4 = 20
Therefore, there are 20 possible combinations of 2 letters.
For the digits, there are 6 options (0, 1, 2, 3, 4, 5) and we need to choose 3 digits without repetition. Again, using the permutation formula:
P(6, 3) = 6! / (6 - 3)! = 6! / 3! = (6 x 5 x 4 x 3!) / 3! = 6 x 5 x 4 = 120
Therefore, there are 120 possible combinations of 3 digits.
To find the total number of possible house numbers, we multiply the number of letter combinations by the number of digit combinations:
Total combinations = 20 x 120 = 2400
If each house number is sold for P100, the total amount of money they will receive is:
2400 x 100 = P240,000
Therefore, they will receive P240,000 if all house numbers are taken.