Hello so I just need some help determining wether my logic is correct or if it's flawed.

Bob starts with a string Superman and then applies one modification to it where a modification consists of inserting a number at some position. This process can produce a string like Super5man.

1. How many different strings can Bob create starting with Superman?

My answer:


So 10 choices and 9 positions: 10 x 90 = 90

Suppose Bob wants to apply two modifications to produce a string like 9Super5man or a string like Super95man.

We want to determine the different strings Bob can create. He does this in two steps:

Step 1. Choose the first number (say 5) and insert it in Superman. (It produces Super5man, etc.)

Step 2. Choose the second number (say 9) and insert it in the string obtained after Step 1. (If the string from step 1 was Super5man, then the result can be 9Super5man or Super95man, etc.)

2. To determine the number of different strings Bob can create, we can just apply the Product Rule. Explain why the product 1 rule will result in overcounting. In particular, what strings will be counted more than once?

My answer

So here i'm at a bit of a loss. I'm not too sure what consist of overcounting here.

Would this be considered over counting

95superman 59superman??

3. How would you solve the problem? Explain why you think it is correct.

My answer:

_ _ s _ _ u _ _ p_ _ e_ _ r_ _ m_ a _ n_ _

20 --> total number of positions

10 ----> total number of digit options

So: (20 x 10) (first option) + (20 x 10) (second option)

  • 2
    $\begingroup$ You have an error in that you are indeed overcounting. Suppose you in the first step choose to insert a $5$ at the front and in the second step choose to insert a $9$ at the second position resulting in 59superman. Compare this to having in the first step chosen to insert a $9$ at the front and in the second step choosing to insert a $5$ in the first position, again resulting in 59superman. For a correct approach, break into cases based on whether both numbers went into the same "slot" between letters, or whether they went into different slots. $\endgroup$ – JMoravitz Mar 30 '17 at 21:49
  • $\begingroup$ Thank you!! That is correct. Wow I would of never thought about that. Now in terms about breaking into these cases how would i go about that? Would it be: (20 x 10) + (19 x 10) Since now only 19 spaces are available? $\endgroup$ – AndrewWallis Mar 30 '17 at 21:56
  • $\begingroup$ As mentioned, break into cases based on whether they went to same slot or different slots. Individually count how many these are and add. For counting if they go to the same slot, first pick which slot it is and then from left-to-right pick which each number is. For counting if they go to different slots simultaneously pick the two slots to be used, and then from left-to-right pick which number was used for each. Apply multiplication principle and addition principle and conclude. (pick simultaneously in case2 to avoid overcounting like we did above) $\endgroup$ – JMoravitz Mar 30 '17 at 22:00

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