Each Sunday the hymns to be sung are displayed on a hymn-board using numbered cards. The cards have the digits 0 to 9 on them. The hymn book has 750 hymns in it. Three different hymns are sung each service.
a) What is the smallest number of cards the church must have to be able to display any combination of 3 hymns?
b) If the number 6 can be used as an upside-down 9, how many cards can be saved from your anwswers to (a)?
the first card: 1-7
the second card: 0-9
the third card: 0-9
I calculate so that is lets see, 28 cards?
YOu can save three two cards, the second set of 9's.
Sorry, I'm still don't understand the suppose answer is a) 66 b) 4
a) To find the smallest number of cards required to display any combination of 3 hymns, we need to consider the maximum possible combinations. Since we have 750 hymns in the hymn book, the maximum number of combinations can be calculated using the concept of combination formula:
C(n, r) = n! / (r!(n-r)!)
In this case, n = 750 (total number of hymns) and r = 3 (number of hymns chosen at a time). Plugging in these values:
C(750, 3) = 750! / (3!(750-3)!)
= 750! / (3! * 747!)
Simplifying further:
C(750, 3) = 750 * 749 * 748 / (3 * 2 * 1)
= 2,736,250
So, there are 2,736,250 possible combinations of 3 hymns from a total of 750 hymns.
To display any of these combinations, we need at least one card for each hymn chosen. Therefore, the smallest number of cards required would be equal to the number of possible combinations:
Smallest number of cards = 2,736,250
b) If we consider the upside-down usage of the number 6 as a 9, we can save a few cards.
Since there are no '9' cards in the set of digits (0-9), we need to make sure that we have enough '6' cards to use as '9' when necessary. The number of '6' cards required depends on the number of combinations that involve '9'.
To find the number of combinations that involve '9', we can calculate the number of combinations when we replace one of the digits with '9'. So, we need to calculate C(750, 2) as now we are choosing 2 hymns out of 750, with one of them being '9'.
C(750, 2) = 750! / (2! * 748!)
Simplifying further:
C(750, 2) = 750 * 749 / (2 * 1)
= 281,625
So, there are 281,625 combinations that involve '9'.
As each combination requires at least one card, and we can use '6' as '9', we need to have 281,625 + 6 = 281,631 cards to display any combination of 3 hymns, along with the upside-down usage of '6' as '9'.
Therefore, the number of cards saved from the answer to part (a) would be:
Number of cards saved = (Smallest number of cards in part (a)) - (Number of cards in part (b))
= 2,736,250 - 281,631
= 2,454,619