A student visits a sports club everyday from monday to friday after scholl hours and plays one of the three games:Cricket,Tennis,Football.In how many ways can he play each of tje 3 games atleast once during a week?
Consider counting from 00000 to 22222 in base 3. There are 3^5 = 243 different values.
We can consider just 81 of these, since however we treat the 0's can also be done for the 1's and 2's.
Of the 81 numbers that have 0's, there is just 1 that is all 0's.
That leaves 80, some of which have no 1's or no 2's.
There are 15 with no 1's, and 15 with no 2's. So, that means there are 50 that have 0,1,2.
Replace 0 with 1 or with 2 and reason the same, so there are 150 different ways to play at least one of the sports during the 5-day week.
I'm sure some combinatorics expert out there that has a better formula.
Well, let's see if we can use a little clown logic to solve this one!
Since the student has to play each of the three games at least once during the week, let's try to figure out the number of ways he can arrange the games.
First, we need to pick one game for each day. We have three choices for the game on Monday, three choices for Tuesday, three choices for Wednesday, and so on.
So, the total number of ways he can pick a game for each day is 3 x 3 x 3 x 3 x 3 = 243.
But wait, we have to take into account that the order of the games doesn't matter. In other words, playing Cricket on Monday and Tennis on Tuesday is the same as playing Tennis on Monday and Cricket on Tuesday.
To find the number of unique combinations, we divide 243 by the number of ways we can arrange each day, which is 5! (5 factorial) since there are 5 weekdays.
Using some clown math, we get 243 / (5 x 4 x 3 x 2 x 1) = 9 ways.
So, there are 9 different ways for the student to play each of the three games at least once during a week. Good luck to that student in mastering all the sports!
To find the number of ways the student can play each of the three games at least once during a week, we can use the principle of inclusion-exclusion.
Let's denote Cricket as C, Tennis as T and Football as F.
Total number of ways to play the games: 3^5 (as the student can play any of the three games on each of the five weekdays)
Now, let's calculate the number of ways when at least one game is not played:
- Let's consider the case when Cricket is not played:
In this case, the student has 2 options for each of the 5 days (either Tennis or Football).
So the number of ways is 2^5.
- Similarly, when Tennis is not played, the number of ways is also 2^5.
- And when Football is not played, the number of ways is 2^5.
However, if we subtract the cases when only one game is not played, we are subtracting them twice. So, we need to add them back once:
- Let's consider the case when both Cricket and Tennis are not played:
In this case, the student has only 1 option for each of the 5 days (only Football).
So the number of ways is 1^5.
- Similarly, when both Cricket and Football are not played, the number of ways is 1^5.
- And when both Tennis and Football are not played, the number of ways is 1^5.
Lastly, if we only consider the case when none of the games are played, we would be subtracting it three times and adding it three times. So, we need to subtract it two times:
- Let's consider the case when none of the games are played:
In this case, the student has no options for each of the 5 days.
So the number of ways is 0^5.
Now, we can calculate the total number of ways when each of the three games is played at least once during a week:
Total number of ways = 3^5 - 2^5 - 2^5 - 2^5 + 1^5 + 1^5 + 1^5 - 0^5
= 243 - 32 - 32 - 32 + 1 + 1 + 1 - 0
= 144
Therefore, the student can play each of the three games at least once during a week in 144 different ways.
To find the number of ways the student can play each of the three games at least once during the week, we can use the principle of inclusion-exclusion.
First, let's count the total number of ways the student can play the games without any restrictions. Since the student can play any of the three games on each day, there are 3 choices for each day. As there are 5 days in a week (Monday to Friday), the total number of ways is 3^5.
Next, let's count the number of ways the student can play only two out of the three games. There are three choices for the excluded game and two choices for each of the remaining games. Since the student can choose which two games to play in (3 choose 2) ways, the total number of ways is 3 * 2^5.
Finally, let's count the number of ways the student can play only one out of the three games. There are three choices for the game to be played and only one choice for each of the remaining days. Since the student can choose which game to play in (3 choose 1) ways, the total number of ways is 3^1 * 1^4.
Now, to find the number of ways the student can play each of the three games at least once during the week, we subtract the cases where the student plays only two games or only one game from the total number of ways:
Number of ways = Total number of ways - Number of ways to play only two games - Number of ways to play only one game
= 3^5 - 3 * 2^5 - 3^1 * 1^4
= 243 - 3 * 32 - 3
= 243 - 96 - 3
= 144
Therefore, there are 144 ways the student can play each of the three games at least once during the week.