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- Question
A school has scheduled three volleyball games, two soccer games, and four basketball games. You have a ticket allowing you to attend three of the games. In how many ways can you go to two basketball games and one of the other events?
Options- A. 25
- B. 30
- C. 50
- D. 75
- Correct Answer
- 30
ExplanationSince order does not matter it is a combination.
The word AND means multiply.
Given 4 basketball, 3 volleyball, 2 soccer.
We want 2 basketball games and 1 other event. There are 5 choices left.
C(n,r)
C(How many do you have, How many do you want)
C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1)
C(4,2) x C(5,1) = (6)(5) = 30
Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.
- 1. Determine the total number of five-card hands that can be drawn from a deck of 52 cards.
Options- A. 2589860
- B. 2598970
- C. 2598960
- D. 2430803 Discuss
- 2. How many ways are there to select a subcommittee of 7 members from among a committee of 17?
Options- A. 19000
- B. 19448
- C. 19821
- D. 19340 Discuss
- 3. Find the number of ways to take 20 objects and arrange them in groups of 5 at a time where order does not matter.?
Options- A. 57090
- B. 15540
- C. 15504
- D. 23670 Discuss
- 4. Find the number of ways to arrange 6 items in groups of 4 at a time where order matters?
Options- A. 720
- B. 640
- C. 740
- D. 360 Discuss
- 5. Find the number of ways to arrange 4 people in groups of 3 at a time where order matters?
Options- A. 20
- B. 16
- C. 24
- D. 36 Discuss
- 6. Find the number of ways to draw a straight, (suit does not matter) beginning with a 4 and ending with a 8?
Options- A. 1024
- B. 1296
- C. 1094
- D. 1200 Discuss
- 7. A certain marathon had 50 people running for first prize, second, and third prize.How many ways are there to correctly guess the first, second, and third place winners?
Options- A. 2
- B. 1
- C. 4
- D. 3 Discuss
- 8. How many number of times will the digit ?7' be written when listing the integers from 1 to 1000?
Options- A. 243
- B. 300
- C. 301
- D. 290 Discuss
- 9. A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?
Options- A. 126
- B. 240
- C. 120
- D. 260 Discuss
- 10. Eight first class and six second class petty officers are on the board of the 56 club. In how many ways can the members elect, from the board, a president, a vice-president, a secretary, and a treasurer if the president and secretary must be first class petty officers and the vice-president and treasurer must be second class petty officers?
Options- A. 1500
- B. 1860
- C. 1680
- D. 1640 Discuss
Permutation and Combination problems
Search Results
Correct Answer: 2598960
Explanation:
When a hand of cards is dealt, the order of the cards does not matter. If you are dealt two kings, it does not matter if the two kings came with the first two cards or the last two cards. Thus cards are combinations. There are 52 cards in a deck and we want to know how many different ways we can put them in groups of five at a time when order does not matter. The combination formula is used.
C(52,5) = 2,598,960
Correct Answer: 19448
Explanation:
Since it does not matter what order the committee members are chosen in, the combination formula is used.
Committees are always a combination unless the problem states that someone like a president has higher hierarchy over another person. If the committee is ordered, then it is a permutation.
C(17,7)= 19,448
Correct Answer: 15504
Explanation:
= 15504
Correct Answer: 360
Explanation:
6P4 = 6! / (6-4)! = 360
Correct Answer: 24
Explanation:
P(4,3)= = 24
Correct Answer: 1024
Explanation:
There are 5 slots.
__ __ __ __ __
The first slot must be a four. There are 4 ways to put a four in the first slot.
There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.
(4)(4)(4)(4)(4) = 1024
Therefore there are 1024 different ways to produce the desired hand of cards.
Correct Answer: 1
Explanation:
There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.
Correct Answer: 300
Explanation:
7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
243 + 54 + 3 = 300
Correct Answer: 126
Explanation:
There are 8 students and the maximum capacity of the cars together is 9.
We may divide the 8 students as follows
Case I: 5 students in the first car and 3 in the second Or
Case II: 4 students in the first car and 4 in the second
Hence, in Case I: 8 students are divided into groups of 5 and 3 in ways.
Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in ways.
Therefore, the total number of ways in which 8 students can travel is
= 56 + 70 = 126.
Correct Answer: 1680
Explanation:
Since two of the eight first class petty officers are to fill two different offices, we write 56
Then, two of the six second class petty officers are to fill two different offices; thus, we write =30
The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders
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