Difference between revisions of "2008 iTest Problems/Problem 67"
Rockmanex3 (talk | contribs) (Solution to Problem 67 — circular seating charts) |
(Added a solution 2 that uses complementary counting) |
||
| Line 6: | Line 6: | ||
Let Alexis sit at a designated spot, which "pins" the circle and prevents any cases from being overcounted due to symmetry. Joshua has <math>4</math> spots to choose from since he can not sit next to Alexis. The remaining five members can sit anywhere, so there are <math>4 \cdot 5! = \boxed{480}</math> distinct ways for the family to sit at the table. | Let Alexis sit at a designated spot, which "pins" the circle and prevents any cases from being overcounted due to symmetry. Joshua has <math>4</math> spots to choose from since he can not sit next to Alexis. The remaining five members can sit anywhere, so there are <math>4 \cdot 5! = \boxed{480}</math> distinct ways for the family to sit at the table. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | We'll use complementary counting. There are <math>6!</math> ways to seat the <math>7</math> members by the same reasoning in the first solution. Now suppose that Alexis and Joshua do sit together. If Alexis is sitting at her designated spot, then there's two ways for Joshua to sit down, and | ||
| + | <math>5!</math> ways for the rest of the family. Then there are <math>6!-2\cdot5!=\boxed{480}</math>. | ||
==See Also== | ==See Also== | ||
Latest revision as of 06:17, 11 March 2019
Contents
Problem
At lunch, the seven members of the Kubik family sits down to eat lunch together at a round table. In how many distinct ways can the family sit at the table if Alexis refuses to sit next to Joshua? (Two arrangements are not considered distinct if one is a rotation of the other.)
Solution
Let Alexis sit at a designated spot, which "pins" the circle and prevents any cases from being overcounted due to symmetry. Joshua has 
spots to choose from since he can not sit next to Alexis. The remaining five members can sit anywhere, so there are 
distinct ways for the family to sit at the table.
Solution 2
We'll use complementary counting. There are 
ways to seat the 
members by the same reasoning in the first solution. Now suppose that Alexis and Joshua do sit together. If Alexis is sitting at her designated spot, then there's two ways for Joshua to sit down, and
ways for the rest of the family. Then there are 
.
See Also
| 2008 iTest (Problems) | ||
| Preceded by: Problem 66 |
Followed by: Problem 68 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100 | ||
