a) A five-digit number ranges from 10000 to 99999. The first digit (leftmost) must be from 1 to 9 (i.e., cannot be 0), while the remaining four digits can be from 0 to 9.
\[ \text{Total numbers} = 9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 90000 \]
For a number to not contain the digit 7, each of its digits must be selected from the set {0,1,2,3,4,5,6,8,9}, which has 9 choices per digit. However, the first digit cannot be 0, so it has only 8 choices.
\[ \text{Numbers without 7} = 8 \cdot 9 \cdot 9 \cdot 9 \cdot 9 = 52488 \]
The numbers that contain at least one digit 7 are found using the complement principle:
\[ \text{Total numbers} - \text{Numbers without 7} \]
\[ 90000 - 52488 = 37512 \]
Closea) The number of ways we can select 4 from 20 is:
\[ C^{20}_4 = 4845 \]
b) The number of ways we can select 2 boys from 12 and 2 girls from 8 is:
\[ C^{12}_2 \text{ and } C^{8}_2 \]
The product of these two will give us the total number of teams:
\[ 66 \times 28 = 1848 \]
c) We can either have a team of only boys, in which case we choose 4 boys from 12:
\[ C^{12}_4 = 495 \]
Or we choose 4 girls from 8:
\[ C^{8}_4 = 70 \]
Thus, the total number of teams is:
\[ 495 + 70 = 565 \]
Closea) The first workstation can be filled in \( C^{12}_4 \) ways. The second workstation can be filled in \( C^{8}_4 \) ways, as 4 people have already been assigned. The final workstation can then be assigned in \( C^{4}_4 \) ways.
We need to find the product of these three cases:
\[ 495 \cdot 70 \cdot 1 = 34650 \]
Closea) Since there are no restrictions on the subjects of the students, we can simply choose 6 students from the total pool of 15 students (10 engineering students and 5 science students).
The number of ways to choose 6 students from 15 is given by the combination formula:
\[ \binom{15}{6} = \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 5005 \]
Thus, the number of possible ways to choose the team is 5005.
b) If only engineering students are chosen, we need to choose all 6 students from the 10 engineering students.
The number of ways to choose 6 engineering students from 10 is:
\[ \binom{10}{6} = \frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} = 210 \]
Thus, the number of possible ways to choose the team with only engineering students is 210.
c) If all five science students are chosen, we must select 5 science students from the 5 available science students. This can only be done in one way since we are choosing all 5.
After choosing all 5 science students, we need to select 1 engineering student from the 10 available engineering students. The number of ways to do this is:
\[ \binom{10}{1} = 10 \]
Thus, the number of possible ways to choose the team with all five science students is 10.
Closea) If the coach and six athletes must sit together, we can treat them as a single unit (or block). This block can be placed in any of the 10 available seats as a single entity.
Since this block occupies 7 consecutive seats, it can be placed in any of the 4 possible positions within the 10-seat row (starting at seat 1, 2, 3, or 4).
Within this block, the 7 individuals can be arranged among themselves in:
\[7! = 5040\]
Thus, the total number of ways to seat them together is:
\[4 \cdot 7! = 4 \cdot 5040 = 20160 \]
b) For this part, the coach is fixed at one end of the row. There are 2 choices for where the coach can sit (either at the leftmost seat or the rightmost seat).
\[ \text{Choices for coach's position} = 2 \]
Once the coach is seated, the remaining 9 seats are available for the 6 athletes. We need to choose 6 seats from the 9 available seats and arrange the athletes in them. This can be done in:
\[ \binom{9}{6} = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} = 84\]
After selecting the 6 seats, the 6 athletes can be arranged in:
\[ 6! = 720 \]
Thus, the total number of ways to arrange the seating is:
\[ 2 \cdot \binom{9}{6} \cdot 6! = 2 \cdot 84 \cdot 720 = 120960 \]
Closea) If Maya finishes immediately after Elena, we can treat Elena and Maya as a pair. This pair must occupy two consecutive positions in the final ranking.
First, we count how many ways the pair can be arranged in the positions:
For the remaining 10 players, they can be arranged in the remaining 10 positions in:
\[10! = 3,628,800 \]
Thus, the total number of ways the 12 players can finish with Maya immediately after Elena is:
\[ 11 \cdot 10! = 11 \cdot 3,628,800 = 39,916,800 \]
b) Now, we want to find the number of ways in which Maya finishes in any position after Elena. This means that Maya can finish anywhere after Elena, but there can be any number of competitors between them.
First, Elena can finish in any of the 12 positions. For each position that Elena occupies, Maya can occupy any of the positions that come after Elena. If Elena is in position \( k \), then Maya can occupy any position from \( k+1 \) to 12, giving us \( 12-k \) possible positions for Maya.
Thus, the total number of ways Maya can finish after Elena is:
\[ \sum_{k=1}^{11} (12-k) = 11 + 10 + 9 + \cdots + 1 = \frac{11 \cdot 12}{2} = 66 \]
After determining the positions for Elena and Maya, we can arrange the remaining 10 players in the remaining 10 positions:
\[ 10! = 3,628,800 \]
Thus, the total number of ways the 12 players can finish with Maya in any position after Elena is:
\[ 66 \cdot 10! = 66 \cdot 3,628,800 = 239,004,800 \]
Closea) We will first calculate the total number of ways to arrange the 15 players without any restrictions, and then subtract the cases where Alex, Ben, Charlie, and David are all standing together.
Step 1: Total Number of Ways Without Any Restrictions
Without any restrictions, the 15 players can be arranged in:
\[ 15! = 15 \times 14 \times 13 \times \cdots \times 1 \]
Step 2: Number of Ways Where Alex, Ben, Charlie, and David Stand Together
Next, we will treat Alex, Ben, Charlie, and David as a single block (since they must not be next to each other). If they are treated as a block, this block can be positioned in any of the 12 available spots among the 15 positions.
So, there are 12 places to position the block.
Within this block, the 4 individuals (Alex, Ben, Charlie, and David) can be arranged in:
\[ 4! = 24 \]
After placing the block, the remaining 11 players can be arranged in the remaining 11 positions in:
\[ 11! = 11 \times 10 \times 9 \times \cdots \times 1 \]
Thus, the total number of ways where Alex, Ben, Charlie, and David are all standing together is:
\[ 12 \times 4! \times 11! = 12 \times 24 \times 11! \]
Step 3: Subtract the Cases Where Alex, Ben, Charlie, and David Are Together
The total number of arrangements without any restrictions is \(15!\), and the number of cases where Alex, Ben, Charlie, and David are together is \(12 \times 4! \times 11!\).
Thus, the number of arrangements where Alex, Ben, Charlie, and David are not standing together is:
\[ 15! - 12 \times 4! \times 11! \]
Now, we calculate the final result:
\[ 15! - 12 \times 4! \times 11! = 15! - 12 \times 24 \times 11! = 1,296,178,329,600 \]
Closea) We can choose any digit from the set \( \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \) for each of the 6 positions in the serial number. Since there are 9 choices for each position, the total number of possible serial numbers is:
\[ 9^6 = 531441 \]
Thus, the total number of possible serial numbers is 531441.
b) (i) We need to choose 6 different digits from the 9 available digits. This can be done in:
\[ 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 60480 \]
Thus, the number of possible serial numbers with different digits is 60480.
b) (ii) The odd digits available are \( \{1, 3, 5, 7, 9\} \), so there are 5 possible digits for each of the 6 positions. Thus, the number of serial numbers that contain only odd digits is:
\[ 5^6 = 15625 \]
Thus, the number of possible serial numbers with only odd digits is 15625.
b) (iii) We can choose any digit from \( \{1, 2, 3, 4, 6, 7, 8, 9\} \), which gives us 8 possible choices for each of the 6 positions. The number of serial numbers that do not contain the digit 5 is:
\[ 8^6 = 262144 \]
Thus, the number of possible serial numbers that do not contain the digit 5 is 262144.
b) (iv) To find the number of serial numbers that contain the digit 5, we can use the complement principle. The total number of serial numbers is \(9^6\), and the number of serial numbers that do not contain the digit 5 is \(8^6\). Therefore, the number of serial numbers that contain the digit 5 is:
\[ 9^6 - 8^6 = 531441 - 262144 = 269297 \]
Thus, the number of possible serial numbers that contain the digit 5 is 269297.
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