Final answer:
To calculate the number of 4-digit decimal codes containing '98' with no repeated digits, we find the available choices for the remaining two slots and consider the possible placements for '98'. There are 8 choices for the first slot and 7 for the second, giving 56 combinations. Factoring in placements for '98', we get a total of 168 codes.
Step-by-step explanation:
The student is asking about the number of possible 4-digit decimal codes that can be formed under specific conditions. To have a code that shows "98" exactly, we assume "98" must be part of the code. Since there should be no repeated digits and "98" are two of the four digits needed, there are two remaining slots to fill with the other digits from 0-9 with the exception of 9 and 8 themselves. We can have any of the remaining numbers (0-7) in these slots.
The number of ways to choose the first remaining digit is 8 (being any number from 0-7). After we have chosen the first digit, there are 7 digits left for the second slot. This results in 8 choices for the first slot and 7 choices for the second slot, which leads to 8 * 7 = 56 different possible codes. However, "98" can appear at the beginning, middle or end of the code, hence we have to consider all the possible placements for "98".
Regarding the placement of "98" in the code, we have the following possibilities:
- "98" at the beginning: XX98
- "98" in the middle: X98X
- "98" at the end: 98XX
There are 3 different placements for "98". So, we must multiply the 56 possible combinations by 3 for each placement option, giving us a total of 56 * 3 = 168 possible 4-digit codes with no repeated digits that include "98".