What you have is the total number of ways to place the caps
Let's actually list them, there are only 6
I will write them as a pair of digits, where the first number is the bottle and the second digit is the cap
11 22 33 ----> all correct
11 23 32 ----> 1 right, 2 wrong
12 21 33 ----> 1 right, 2 wrong
12 23 31 ----> all wrong
13 21 32 ---- all wrong
13 22 31 ----> 1 right, 2 wrong
notice, the case of 2 right, 1 wrong is not possible
Three bottles have three different colored caps.How many ways can a cap be put on each bottle such that each one is wrong?
What i did here is,
3 x 2 x 1 = 6
1 answer