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- Question
A school committee of 5 is to be formed from 12 students.How many committees can be formed if John must be on the committee?
Options- A. 11P4
- B. 11C4
- C. 11P5
- D. 11C5
- Correct Answer
- 11C4
ExplanationIf John must be on the committee,we have 11 students remaining,out of which we choose 4. So,
- 1. How many ways can you select 17 songs for mix CD out of possible 38 songs?
Options- A. 2878
- B. 2878 x 10^2
- C. 290183753
- D. 2878 x 10^10 Discuss
- 2. A pizza can have 3 toppings out of possible 7 toppings.How many different pizza's can be made?
Options- A. 49
- B. 35
- C. 27
- D. 25 Discuss
- 3. A committee of 4 people is to be formed from a group of 9 people.How many possible committees can be formed?
Options- A. 120
- B. 162
- C. 126
- D. 170 Discuss
- 4. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
Options- A. 720
- B. 360
- C. 120
- D. 640 Discuss
- 5. In how many ways, can zero or more letters be selected form the letters AAAAA?
Options- A. 5
- B. 6
- C. 2
- D. 8 Discuss
- 6. From a deck of 52 cards, a 7 card hand is dealt.How many distinct hands are there if the hand must contain 2 spades and 3 diamonds ?
Options- A. 7250100
- B. 7690030
- C. 7250000
- D. 3454290 Discuss
- 7. From a deck of 52 cards, a 5 card hand is dealt.How may distinct five card hands are there if the queen of spades and the four of diamonds must be in the hand?
Options- A. 52C5
- B. 50C3
- C. 52C4
- D. 50C4 Discuss
- 8. In how many ways can you choose one or more of 12 different candies?
Options- A. 4054
- B. 4050
- C. 4095
- D. 4059 Discuss
- 9. How many 3 letters words can be formed using the letters of the words hexagon?
Options- A. 120
- B. 210
- C. 160
- D. 200 Discuss
- 10. In how many different ways can 3 students be associated with 4 chartered accountants,assuming that each chartered accountant can take at most one student?
Options- A. 12
- B. 36
- C. 24
- D. 16 Discuss
Permutation and Combination problems
Search Results
Correct Answer: 2878 x 10^10
Explanation:
since it is a combination = = 2878 x 10^10
Correct Answer: 35
Explanation:
There are 7 toppings in total,and by selecting 3,we will make different types of pizza.
This question is a combination since having a different order of toppings will not make a different pizza.
So, =35
Correct Answer: 126
Explanation:
This question is a combination since order is not important.
Answer = = 126
Correct Answer: 360
Explanation:
Two horses A and B, in a race of 6 horses... A has to finish before B
if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!
if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!
if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!
if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4!
if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..
A cannot finish 6th, since he has to be ahead of B
Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360
Correct Answer: 6
Explanation:
Selecting zero'A's= 1
Selecting one 'A's = 1
Selecting two 'A's = 1
Selecting three 'A's = 1
Selecting four 'A's = 1
Selecting five 'A's = 1
=> Required number ofways =6
Correct Answer: 7250100
Explanation:
There are 13 spades,we must include 2: 13
There are 13 diamonds,we must include 3:
Since we can't have more than 2 spades and 3 diamonds,the remaining 2 cards must be pulled out from the 26 remaining clubs and hearts :
Therefore, = 7250100
Correct Answer: 50C3
Explanation:
If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.
So, 50
Correct Answer: 4095
Explanation:
Each candy can be dealt with in two ways.It can be chosen or not chosen.This will give 2 possibilites for the first candy, 2 for the second, and so on.by multiplying the cases together we get . Since the case of no candy being selected is not an option, we have to subtract 1 from our answer.
Therefore,there are - 1 = 4095 ways of selecting one or more candies
Correct Answer: 210
Explanation:
Since the word hexagon contains 7 different letters,the number of permutations is = 7 x 6 x 5 =210
Correct Answer: 24
Explanation:
Number of permutations = = 24
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