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- Question
There are 7 men and 10 women on a committee selection pool.A committee consisting of President,Vice-President, and Treasurer is to be formed.How many ways can exactly two men be on the committe ?
Options- A. 1200
- B. 1240
- C. 1260
- D. 1620
- Correct Answer
- 1260
ExplanationThere are ways of selecting two men, and ways of selecting a woman.Since each position in the committee is different,arrange the three people in 3! ways.
So,
- 1. A research team of 6 people is to be formed from 10 chemists,5 politicians, 8 economists and 15 biologists.How many teams have atleast 5 chemists?
Options- A. 7350
- B. 6400
- C. 6379
- D. 7266 Discuss
- 2. How many arrangements of the word TRIGONAL can be made if only two vowels and three consonants are used?
Options- A. 6300
- B. 3600
- C. 6400
- D. 7200 Discuss
- 3. A student council of 5 members is to be formed from a selection pool of 6 boys and 8 girls.How many councils can have Jason on the council?
Options- A. 715
- B. 725
- C. 419
- D. 341 Discuss
- 4. From a deck of 52 cards, a 5 card hand is dealt.How many distinct hands can be formed if there are atleast 2 queens?
Options- A. 103336
- B. 120000
- C. 108336
- D. 108333 Discuss
- 5. There are fourteen juniors and twenty-three seniors in the Service Club. The club is to send four representatives to the State Conference. If the members of the club decide to send two juniors and two seniors, how many different groupings are possible ?
Options- A. 23024
- B. 24023
- C. 23023
- D. 25690 Discuss
- 6. If each of the 8 teams in a league must play each other three times,how many games will be played?
Options- A. 72
- B. 84
- C. 36
- D. 79 Discuss
- 7. In how many different ways can five persons stand in a line for a group photograph?
Options- A. 120
- B. 240
- C. 360
- D. 720 Discuss
- 8. First,second and third prizes are to be awarded at an engineering fair in which 13 exhibits have been entered.In how many different ways can the prizes be awarded?
Options- A. 1736
- B. 1716
- C. 1216
- D. 1346 Discuss
- 9. When John arrives in New York,he has eight stops to see, but he has time only to visit six of them.In how many different ways can he arrange his schedule in New York?
Options- A. 20610
- B. 24000
- C. 20160
- D. 21000 Discuss
- 10. A local delivery company has three packages to deliver to three different homes. if the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?
Options- A. 3
- B. 5
- C. 3!
- D. 5! Discuss
Permutation and Combination problems
Search Results
Correct Answer: 7266
Explanation:
Correct Answer: 3600
Explanation:
First we need to choose two vowels and then three consonants . Now that we have 5 letters required to make the word,arrange them in 5! ways.
So,
Correct Answer: 715
Explanation:
If Jason is on th ecouncil,this reduces the selction pool to only 13 people,out of which we still need to select 4.
So, = 715
Correct Answer: 108336
Explanation:
The total possible cases would be a 5 card hand with no restrictions : 5
The unwanted cases are:
no queens(out of 48 non-queens cards we get 5)
only 1 queen(out of 4 queens we get 1,and out of 48 non-queens we get 4)
Therefore,
Correct Answer: 23023
Explanation:
Choose 2 juniors and 2 seniors.
Correct Answer: 84
Explanation:
Since,each member of the league must meet every other member of the league.If they only played each other once,there would be games.Since,each pairing of teams will occur three times,the answer will be triple.
Therfore, =84
Correct Answer: 120
Explanation:
This is the number of permutations of five things taken all at a time.
Therefore, answer = = 120 ways
Correct Answer: 1716
Explanation:
= 13!/10! = 1716
Correct Answer: 20160
Explanation:
He can arrange his schedule in = 20160 ways
Correct Answer: 5
Explanation:
The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are:
One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.
We can calculate each of these cases and then add them together, or approach this problem from a different angle.
The only case which is left out of the condition is the case where no wrong packages are delivered.
If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.
There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package.
The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
1=1
Determine the total number of ways the three packages can be delivered.
3=6
The number of ways at least one house gets the wrong package is:
6?1=5
Therefore there are 5 ways for at least one house to get the wrong package.
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