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- Question
Find the number of ways to take 4 people and place them in groups of 3 at a time where order does not matter?
Options- A. 4
- B. 12
- C. 36
- D. 16
- Correct Answer
- 4
ExplanationSince order does not matter, use the combination formula
= 24/6 = 4
- 1. 12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?
Options- A. 500
- B. 490
- C. 495
- D. 540 Discuss
- 2. Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct
Options- A. 65000
- B. 64000
- C. 72000
- D. 36000 Discuss
- 3. In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
Options- A. 120960
- B. 120000
- C. 146700
- D. None of these Discuss
- 4. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Options- A. 11670
- B. 12000
- C. 11760
- D. 20050 Discuss
- 5. Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel ?from A to E?
Options- A. 36
- B. 64
- C. 74
- D. 72 Discuss
- 6. How many ways are there to deal a five-card hand consisting of three eight's and two sevens?
Options- A. 36
- B. 72
- C. 24
- D. 16 Discuss
- 7. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
Options- A. 4!/2!
- B. 3!/2!
- C. (4! x 3!) / 2!
- D. 36 Discuss
- 8. In how many ways can the letters of the word "PROBLEM" be rearranged to make 7 letter words such that none of the letters repeat?
Options- A. 49
- B. 7!
- C. 7^7
- D. 7^3 Discuss
- 9. 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
- 10. 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
Permutation and Combination problems
Search Results
Correct Answer: 495
Explanation:
For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is = 495.
Therefore, we can draw 495 quadrilaterals
Correct Answer: 65000
Explanation:
Out of 26 alphabets two distinct letters can be chosen in ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways.
Combined with letters there are X 100 ways = 65000 ways to choose vehicle numbers.
Correct Answer: 120960
Explanation:
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8!/(2! x 2!)= 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters =4!/2!= 12.
Required number of words = (10080 x 12) = 120960
Correct Answer: 11760
Explanation:
Required number of ways= = ( )= 11760.
Correct Answer: 72
Explanation:
The bus fromA to B can be selected in 3 ways.
The bus from B to C can be selected in 4 ways.
The bus from C toD can be selected in 2 ways.
The bus fromD to E can be selected in 3 ways.
So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72
Correct Answer: 24
Explanation:
If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed.
The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards.
We have 4 eights and 4 sevens.
We want 3 eights and 2 sevens.
C(have 4 eights, want 3 eights) x C(have 4 sevens, want 2 sevens)
C(4,3) x C(4,2) = 24
Therefore there are 24 different ways in which to deal the desired hand.
Correct Answer: (4! x 3!) / 2!
Explanation:
ABACUS is a 6 letter word with 3 of the letters being vowels.
If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.
These 4 elements can be rearranged in 4! Ways.
The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.
Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!
Correct Answer: 7!
Explanation:
There are seven positions to be filled.
The first position can be filled using any of the 7 letters contained in PROBLEM.
The second position can be filled by the remaining 6 letters as the letters should not repeat.
The third position can be filled by the remaining 5 letters only and so on.
Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.
Correct Answer: 23023
Explanation:
Choose 2 juniors and 2 seniors.
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,
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