There are 15 folks within the class, and the variety of mates can solely go from 0 to 14, and if there may be somebody with 0 mates, there will probably be nobody with 14 mates. So at most there are 14 distinct variety of mates, and it’s both 0-13 or 1-14. And word that if Tiffany’s 14 classmates have 0-13 quantity mates, we are able to change that to 1-14 just by making the one with 0 pal to be mates with those that had 7-13 mates (making these to have 8-14 mates, and the one who had 0 pal to have 7 mates, not altering these with 1-6 mates, so we now have 1-14). Since this does not have an effect on the variety of mates Tiffany has, we are able to assume that it’s the case of 1-14 variety of mates that’s occurring right here (the identical reasoning beneath can work with 0-13).
Let the 14 mates be referred to as C1, C2, …, C14, in line with the variety of mates they’ve. So C1 has 1 pal, C2 has 2 mates, and so forth. So we now have Tiffany + C1-C14 (ought to we name it “Tiffany and the 14 C’s”?)
Now discover that:
- The one with 14 mates (C14) needs to be mates with everybody else, together with Tiffany. Now this makes the one with just one pal (C1) not pal with Tiffany, since C1 is already pal with C14 and C1 solely has one pal. From right here we all know that Tiffany is pal with C14, and never with C1.
- The one with 13 mates (C13) needs to be mates with everybody else besides C1. Now this makes the one with two mates (C2) not pal with Tiffany, since C2 is already mates with C14 and C13. From right here we all know that Tiffany is pal with C13, and never with C2.
- Continuing the sample of pairing off C(N) with C(15-N), we now have C14, C13, C12, C11, C10, C9, and C8 as Tiffany’s mates. And the remainder (C1, C2, …, C7) should not mates with Tiffany.
Tiffany has 7 mates.