creative math question - help! beyond my skills!

Question

Bedtime Crumblies is testing a recipe for a new line of snack crackers.30 people each sample four recipes labeled A,B,C and D.

and rank them 1st, 2nd, 3rd, and 4th in order of taste preference. Recipe D received no 1st place votes; recipe A received no 3rd place votes;all preferred B to C; 1/6th preferred D to A; one half preferred B to A; 1/5 preferred D to B; only two put B in last place.

 

A. list all possible rankings of the 30 tasters - show work

 

B. How many tasters recorded each of the possible rankings if no two received the same number of votes? how did you get this answer?

Answer 1

OOPS, sorry, there is a solution - at least to part A.  

 

You need to write down all 24 possible rankings and then eliminate the ones that don't satisfy your criteria.  The 24 possible rankings are 

 

ABCD   ABDC   ACBD   ACDB   ADBC   ADCB 

BACD   BADC   BCAD   BCDA   BDAC   BDCA

CABD   CADB   CBAD   CBDA   CDAB   CDBA

DABC   DACB   DBAC   DBCA   DCAB   DCBA

 

D received no 1st place votes so you can cross off the entire 4th row.  

All preferred B to C, so you can cross off the entire 3rd row as well as ACBD, ACDB and ADCB

A received no 3rd place votes so you can cross off BCAD and BDAC 

 

This eliminates 17 potential rankings, so we are left with the following 7 possible rankings 

 

    ABCD   ABDC   ADBC

BACD  BADC  BCDA  BDCA

 

Now for part B.

 

1/5 preferred D to B.  ADBC is the only possible ranking which prefers D to B, so it must have been recorded by 6 of the 30 tasters.  

 

2 put D in last place, which are rankings ABCD or BACD.  These did not receive 1 vote each, so one of them received 2 votes and the other none.  

 

1/6 preferred D to A.  There are 2 rankings which prefer D to A, namely BCDA and BDCA.  5 of the 30 tasters picked one of these rankings.  5 = 0 + 5 or 1 + 4 or 2 + 3 votes.  Since 0 and 2 are already used for the D in last place rankings, one of BCDA and BDCA must have received 1 vote and the other 4 votes.  

 

One half preferred B to A.  This implies that ABCD + ABDC + ADBC = 15.  We know that ABCD is 0 or 2, and that ADBC is 6, so ABDC must be 7 or 9

 

BACD + BADC + BCDA + BDCA must also be 15.  We know that BACD is 0 or 2, and that the sum of BCDA and BDCA is 5, so BADC must be 8 or 10  

 

So we have the following possibilities for the number of tasters recording each ranking 

ABCD = 0, ABDC = 9, ADBC = 6, BACD = 2, BADC = 8, BCDA = 1, BDCA = 4

ABCD = 2, ABDC = 7, ADBC = 6, BACD = 0, BADC = 10, BCDA = 1, BDCA = 4

 

Note that the 1 and 4 and BCDA and BDCA could also be switched in each line 

 

In the 1st case the number of tasters recording each possible ranking is 

0, 1, 2, 4, 6, 8 and 9

In the 2nd case it is 

0, 1, 2, 4, 6, 7 and 10

 

Both cases seem to satisfy all the criteria, so I don't see any way to distinguish between them