a magician has a repertoire of 8 magic tricks of which he performs 2 on each show. In how many ways can he plan his program for 3 successive shows if:
he does not want to perform the same trick on any 2 successive shows?
He wants to use the same trick for the opening trick of each show?
Can you please show me how to set it up?
Consider the cases of show 1 and then show 2 and then show 3
For show one he has ____ x _____ (8 tricks to pick first then 7 for his second trick (so the two tricks are not the same).
Then we ADD on case 2 where her performs 2 more tricks. Now at show two he wants the exact same first number as in show 1
so
_____ x _____
becomes 1 (for the exact first number from show 1) then ____ (but his second number can not be any of the 7 he used as trick 2 in the first show)
so ____ x _____ becomes 1 x6
So for shows 1 and 2 we have
8x7 + 1x6
with this logic can you generate the third show?
To solve this problem, we can use combinations and permutations. Let's go through each case step by step:
1. If the magician does not want to perform the same trick on any two successive shows:
In this case, for the first show, the magician has 8 tricks to choose from. For the second show, since he cannot perform the same trick as the first show, he will have 7 remaining tricks to choose from. Similarly, for the third show, he will have 7 tricks to choose from again (since he cannot perform the trick from the second show).
To calculate the total number of ways, we multiply the number of choices for each show: 8 * 7 * 7 = 392 ways.
2. If the magician wants to use the same trick for the opening trick of each show:
In this case, since the magician wants to use the same trick for the opening trick, he has only 8 options to choose from for the first show.
For the second show, since he cannot perform the same trick as the first show, he will have 7 remaining tricks to choose from.
For the third show, since the magician wants to use the same trick for the opening trick again, he will have only 1 option left.
To calculate the total number of ways, we multiply the number of choices for each show: 8 * 7 * 1 = 56 ways.
So, the magician can plan his program for 3 successive shows in 392 ways if he does not want to perform the same trick on any two successive shows, and in 56 ways if he wants to use the same trick for the opening trick of each show.