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- Question
How many 3 letters words can be formed using the letters of the words hexagon?
Options- A. 120
- B. 210
- C. 160
- D. 200
- Correct Answer
- 210
ExplanationSince the word hexagon contains 7 different letters,the number of permutations is = 7 x 6 x 5 =210
- 1. In how many ways can you choose one or more of 12 different candies?
Options- A. 4054
- B. 4050
- C. 4095
- D. 4059 Discuss
- 2. 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
- 3. 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
- 4. 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 Discuss
- 5. 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
- 6. 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
- 7. When Dr. Ram arrives in his dispensary, he finds 12 patients waiting to see him. If he can see only one patients at a time,find the number of ways, he can schedule his patients if 3 leave in disgust before Dr. Ram gets around to seeing them.
Options- A. 479001600
- B. 79833600
- C. 34879012
- D. 67800983 Discuss
- 8. In how many ways can 4 sit at a round table for a group discussions?
Options- A. 9
- B. 3
- C. 6
- D. 12 Discuss
- 9. In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
Options- A. 15
- B. 10
- C. 25
- D. 20 Discuss
- 10. In how many different ways can 6 different balls be distributed to 4 different boxes, when each box can hold any number of ball?
Options- A. 2048
- B. 1296
- C. 4096
- D. 576 Discuss
Permutation and Combination problems
Search Results
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: 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: 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: 11C4
Explanation:
If John must be on the committee,we have 11 students remaining,out of which we choose 4. So,
Correct Answer: 2878 x 10^10
Explanation:
since it is a combination = = 2878 x 10^10
Correct Answer: 24
Explanation:
Number of permutations = = 24
Correct Answer: 79833600
Explanation:
There are 12 - 3 = 9 patients.They can be seen in = 79833600 ways
Correct Answer: 6
Explanation:
We get this using the formula of circular permutations i.e., (4 - 1)! = 3! =6
Correct Answer: 25
Explanation:
The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1
Case a. Number of ways of achieving the first option 2?2?1
Two toys out of the 5 can be selected in
5 ways. Another 2 out of the remaining 3 can be selected in
ways and the last toy can be selected in
way.
However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.
Therefore, total number of ways of achieving the 2?2?1 option is:
ways.
Case b. Number of ways of achieving the second option 3?1?1
Three toys out of the 5 can be selected in
ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3?1?1 option is
=10 ways.
Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways
Correct Answer: 4096
Explanation:
Every ball can be distributed in 4 ways.
Hence the required number of ways = 4 x 4 x 4 x 4 x 4 x 4 = 4096
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