Could someone please help me out with the data homework? All I need is someone to check my answers. If any are incorrect could you please explain what I did wrong.
1. A student must answer 7 of 10 questions on an exam. How many different sets of
questions can be answered if
a) there are no restrictions?
b) four of the first five questions must be answered?
a) 10C7 = 120 b) 5C4 x 5C3 = 50
2. A pizza can be ordered with any number of 8 different toppings OR with no toppings at
all.
a) Find the number of 4 topping orders available.
b) Find the total number of possible orders.
a) 8C4 = 70 b) 2^8 – 1 = 255
3. How many five-digit odd numbers can be formed from the digits of the number 5 390
462?
7C5 = 21
21 x 3/7
= 9
4. In the game of poker, each player is dealt a hand of five cards. How many hands contain
a) all hearts?
b) exactly 2 aces?
c) at least one diamond?
d) a full house?
a) 13C5 = 1287
b) 4C2 x 48C3 = 103 776
c) final answer = 2023203
d) 52C3 x 49C2 =
5. A club has 25 members. In how many ways can
a) a committee of 3 members be chosen?
b) the offices of president, vice-president, secretary, and treasurer be filled?
c) the members from (b) stand in line for a picture if the president and vicepresident
must stand side by side?
a) 25C3 = 2300
b) 25P3 = 13 800
c) 2(2!) = 4
6. In how many ways can 8 boys sit in a row if
a) there are no restrictions?
b) Jimmy and Bobby must be kept apart?
a) 8! = 40 320
b) 8! - 2(7!) = 30 240
Thank you so much
I am sorry I didn't realize it cut off
b) 4C2 x 48C3 = 103 776
c) final answer = 2023203
d) 52C3 x 49C2 = 5.
A club has 25 members. In how many ways can
a) a committee of 3 members be chosen?
b) the offices of president, vice-president, secretary, and treasurer be filled?
c) the members from (b) stand in line for a picture if the president and vicepresident
must stand side by side?
a) 25C3 = 2300
b) 25P3 = 13 800
c) 2(2!) = 4
6. In how many ways can 8 boys sit in a row if
a) there are no restrictions?
b) Jimmy and Bobby must be kept apart?
a) 8! = 40 320
b) 8! - 2(7!) = 30 240
Sure, I can help you with your data homework. Let's go through each question one by one.
1. For the first question, we need to find the number of different sets of questions that can be answered when there are no restrictions and when four of the first five questions must be answered.
a) To find the number of different sets of questions when there are no restrictions, we can use the combination formula. The formula for combinations is nCr, where n is the total number of questions (10) and r is the number of questions to be answered (7 in this case). So, the answer is 10C7 = 120.
b) To find the number of different sets of questions when four of the first five questions must be answered, we apply the combination formula for both sets of questions. The first set would be choosing 4 questions out of 5, which is 5C4 = 5. The second set would be choosing 3 questions out of the remaining 5 questions, which is 5C3 = 10. Now, we multiply these two combinations together, giving us 5C4 x 5C3 = 50.
2. For the second question, we need to find the number of 4-topping pizza orders available and the total number of possible orders.
a) To find the number of 4-topping pizza orders available, we can use the combination formula. The formula for combinations is nCr, where n is the total number of toppings (8) and r is the number of toppings to be chosen (4 in this case). So, the answer is 8C4 = 70.
b) To find the total number of possible orders, we need to consider two possibilities: ordering with toppings and ordering with no toppings at all. We subtract 1 from 2^8 because 1 possibility is ordering with no toppings. So, the answer is 2^8 - 1 = 255.
3. For the third question, we need to find the number of five-digit odd numbers that can be formed from the digits of the number 5 390 462.
To solve this problem, we can break it down into steps:
- First, we need to choose 5 digits out of the 7 given digits. This can be calculated using the combination formula. So, the answer is 7C5 = 21.
- Next, we need to consider that the units digit must be odd. Out of the 5 chosen digits, 3 are odd (5, 3, and 9) and 2 are even (0 and 2). Therefore, the probability of choosing an odd number for the units digit is 3/7.
- Finally, we multiply the two results together: 21 x 3/7 = 9.
4. For the fourth question, we need to find the number of poker hands that contain all hearts, exactly 2 aces, at least one diamond, and a full house.
a) To find the number of hands that contain all hearts, we choose 5 cards out of the 13 hearts in the deck. So, the answer is 13C5 = 1287.
b) To find the number of hands that contain exactly 2 aces, we choose 2 aces out of the 4 in the deck and 3 non-aces out of the remaining 48 cards. So, the answer is 4C2 x 48C3 = 103,776.
c) To find the number of hands that contain at least one diamond, we need to calculate 1 - (the probability of all cards being non-diamonds). The probability of choosing a non-diamond card is 39/52, and since we have 5 cards in a hand, we can calculate (39/52)^5. Thus, the answer is 1 - (39/52)^5.
d) To find the number of hands that contain a full house, we choose 3 cards out of the 13 possible ranks (e.g., three 2s, three Queens) and 2 cards out of the remaining 12 ranks. So, the answer is 13C3 x 12C2.
5. For the fifth question, we need to find the number of ways a committee of 3 members can be chosen, the number of ways to fill the offices of president, vice-president, secretary, and treasurer, and the number of ways the members from part (b) can stand in line for a picture if the president and vice-president must stand side by side.
a) To find the number of ways a committee of 3 members can be chosen, we can use the combination formula. The answer is 25C3 = 2300.
b) To find the number of ways to fill the offices, we can use the permutation formula. The answer is 25P3 = 13,800.
c) To find the number of ways the members can stand in line if the president and vice-president must stand side by side, we can first consider the president and vice-president as one unit (e.g., PV). This leaves us with three units (PV, S, T) in which the remaining members can stand. Since the president and vice-president can switch positions, we multiply the number of positions by 2. So, the answer is 2(2!) = 4.
6. For the sixth question, we need to find the number of ways 8 boys can sit in a row, with no restrictions and with Jimmy and Bobby kept apart.
a) To find the number of ways with no restrictions, we can use the factorial formula. The answer is 8! = 40,320.
b) To find the number of ways with Jimmy and Bobby kept apart, we consider Jimmy and Bobby as one unit (e.g., JB). This leaves us with 7 units (JB, A, B, C, D, E, F) in which the remaining boys can sit. Since JB can switch positions, we multiply the number of positions by 2. So, the answer is 8! - 2(7!) = 30,240.
I hope this helps! Let me know if you have any further questions.