Method 1: We place the consonants first.
Since ALGEBRA has seven letters, we have seven positions to fill. If we place the B, G, L, and R in that order, we have seven choices where to place the B, six choices where to place the G, five choices where to place the L, and four choices where to place the R. Once the consonants have been placed, there is only one way to arrange the vowels in the remaining positions so that the E appears between the two A's. Hence, there are $$7 \cdot 6 \cdot 5 \cdot 4$$ ways of arranging the letters of the word ALGEBRA so that the relative positions of the vowels are preserved.
Method 2: We place the vowels first.
We choose three of the seven positions for the three vowels. There is only one way to place the vowels in these positions that preserves their relative order. This leaves four positions. We can arrange the four consonants in these positions in $4!$ ways. Hence, the number of ways of arranging the letters of the word ALGEBRA that preserves the relative order of the vowels is $$\binom{7}{3} \cdot 4!$$ Note that this is essentially a rephrasing of @Joffan's solution.
Method 3: We use symmetry.
First, we count the total number of distinguishable arrangements of the letters of the word ALGEBRA. We choose two of the seven positions for the A's. Once the A's have been placed, we can arrange the remaining five distinct letters in the remaining five positions in $5!$ ways. Hence, the number of distinguishable arrangements of the letters of the word ALGEBRA is $$\binom{7}{2} \cdot 5!$$ By symmetry, in one third of these arrangements does the E appear somewhere between the two A's. Hence, the number of arrangements of the word ALGEBRA in which the relative order of the vowels is preserved is $$\frac{1}{3} \binom{7}{2} \cdot 5! = 840$$
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A permutation is a collection or a combination of objects from a set where the order or the arrangement of the chosen objects does matter. In other words, a permutation is an arrangement of objects in a definite order, For example, if we have two elements A and B, then there are two possible arrangements, ( A B ) and ( B A ).
- nPn = n(n – 1) (n -2)… 3x2x1=n!
- nP0 = 1
- nP1 = n
- nPn-1 = n!
- nPr = n.n-1Pr-1 = n(n-1)n-2Pr-2
The number of permutations when ‘r‘ elements are arranged out of a total of ‘n’ elements is n Pr = n! / (n – r)! For example, let n = 4 (A, B, C and D) and r = 2 (All permutations of size 2). The answer is 4!/(4-2)! = 12. The twelve permutations are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, and DC.
A Combination is the different selections of a given number of elements taken one by one, or some, or all at a time. For example, if we have two elements A and B, then there is only one way select two items, we select both of them.
Key Point
- nCr is a natural number
- nC0=(nCn)=1
- nC1=n
- nCr=(nCn−r)
- nCx=nCy ⇒x=y or x+y=n
- n.n−1Cr−1=(n−r+1)×nCr−1
The number of combinations when ‘r’ elements are selected out of a total of ‘n’ elements is n C r = n! / ((r !) x (n – r)!). For example, let n = 4 (A, B, C and D) and r = 2 (All combinations of size 2). The answer is 4!/((4-2)!*2!) = 6. The six combinations are AB, AC, AD, BC, BD, and CD.
Note: In the same example, we have different cases for permutation and combination. For permutation, AB and BA are two different things but for selection, AB and BA are the same.
Sample Problems
Question 1: How many words can be formed by using 3 letters from the word “DELHI”?
Solution: The word “DELHI” has 5 different words. Therefore, required number of words = 5 P 3 = 5! / (5 – 3)!
Required number of words = 5! / 2! = 120 / 2 = 60
Question 2: How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are always together?
Solution: In these types of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character. So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times. => Number of possible arrangements = 5! / 2! = 60 Now, the two vowels can be arranged in 2! = 2 ways. => Total number of possible words such that the vowels are always together= 60 x 2 = 120
Question 3: In how many ways, can we select a team of 4 students from a given choice of 15?
Solution : Number of possible ways of selection = 15 C 4 = 15 ! / ((4 !) x (11 !))
Number of possible ways of selection = (15 x 14 x 13 x 12) / (4 x 3 x 2 x 1) = 1365
Question 4: In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 boys and 2 girls out of 5 girls?
Solution : Number of ways 3 boys can be selected out of 6 = 6 C 3 = 6 ! / [(3 !) x (3 !)] = (6 x 5 x 4) / (3 x 2 x 1) = 20 Number of ways 2 girls can be selected out of 5 = 5 C 2 = 5 ! / [(2 !) x (3 !)] = (5 x 4) / (2 x 1) = 10 Therefore, total number of ways of forming the group = 20 x 10 = 200
Question 5: How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are never together?
Solution: we assume all the vowels to be a single character, i.e., “IE” is a single character. So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times. => Number of possible arrangements = 5! / 2! = 60 Now, the two vowels can be arranged in 2! = 2 ways. => Total number of possible words such that the vowels are always together = 60 x 2 = 120 ,
total number of possible words = 6! / 2! = 720 / 2 = 360 Therefore, total number of possible words such that the vowels are never together 240
Permutation Combination Questions: Solved 27 Permutation Combination Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.
1. An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …., 9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can however potentially create confusion, when read upside down-for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?
SHOW ANSWER
Correct Ans:71
Explanation:
The available digits are 0,1,2, …9. The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digits repetition not allowed). Thus, the code can be made in 9 × 9 = 81 ways. Now there are only 4 digits 1, 6, 8, 9 which can create confusion. Hence, the total number of codes which create confusion are = 4 × 3 = 12. Out of these 12 codes 69 and 96 will not create confusion. Hence, in total 12 – 2 = 10 codes will create confusion.
Hence, the total codes without confusion are 81 – 10 = 71.
2. If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that at least 1 blue can and 1 red can remains in the refrigerator.
SHOW ANSWER
Correct Ans:455
Explanation:
Given
Total No of Cans = 12
Total No of Red Cans = 7
Total No of Blue Cans = 5
Given Condition - No of Possible ways to draw 8 balls from the refrigerator which contains at least 1 blue and 1 red can
No of Possible ways are => (6,2) (5,3) (4,4) => 7 C 6 * 5 C2 => 7 * 10 => 70 => 7C5 * 5 C3 => 7 * 3 * 5 * 2 => 21 * 10 => 210 => 7C4 * 5C4 => 35 * 5 => 175 Total Ways = 70 +210 + 175=> 455.
3. There is meeting of 20 delegates that is to be held in a hotel. In how many ways these delegates can be seated along a round table, if three particular delegates always seat together?
SHOW ANSWER
Correct Ans:17! 3!
Explanation:
Give Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17! And now that three can be arranged in 3! Ways.
So, 17! 3! is the correct answer.
4. In how many ways a four digit even number can be formed by using the digits 2,3,5,8 exactly once?
SHOW ANSWER
Correct Ans:12
Explanation:
Solution is
Given
Four digit even number can be formed by using the digits 2,3,5,8
Since the number has to be an even number,the unit digit has to be 2 or 8.
First three places can be filled by remaining three digits.
&nb
5. In a plane there are totally 8 points (no three points are collinear), how many lines can be drawn ?
SHOW ANSWER
Correct Ans:28
Explanation:
To draw a line we need 2 points. From the given 8 points , we need to choose 2 points Number of ways in which this can be done is 18C2= (8 x 7 ) / 2 = 28
6. How many 4 digit even number can formed by using the digits 1,3,7 and 8 only once ?
SHOW ANSWER
Correct Ans:6
Explanation:
We can form 24 numbers with digits 1,3,7 and 8 using only once. To be an even number, the units digit has to even number, in this case only 8 should come in the unit's place . The remaining places, 3 positions can be filled in 3! , that is 6 ways
7. How many words can be formed with or without meaning by taking all the letters from the word TAKEN ?
SHOW ANSWER
Correct Ans:120
Explanation:
There are totally 4 letters in the word TAKEN T, A, K, E and N. No. Of Words = 5 x 4 x 3 x 2 x 1 = 120
8. In how many different ways can the letters of the word 'OFFICES' be arranged ?
SHOW ANSWER
Correct Ans:2520
Explanation:
The word OFFICES contains O,F, F I, C, E, S contains 7 letters, in which 2 are identical. No. Of words that can formed = 7 ! / 2 ! = 2520
SHOW ANSWER
Correct Ans:600
Explanation:
25! / 23! = 23! X 24 x 25 / ( 23!) = 24 x 25 = 600
10. In a party there were totally 20 people, each person shook his hands with the other person. How many hand shakes would have taken place ?
SHOW ANSWER
Correct Ans:190
Explanation:
There are 20 people. Every person has to shake hands with the other person. Which means we have to find the number of ways of choosing 2 people from the 20. The number of ways it ca happen = 20C2 = ( 20 x 19 ) / (1 x 2) = 190
11. A letter lock consists of 4 rings, each ring contains 9 non-zero digits. This lock can be opened by setting a 4 digit code with the proper combination of each of the 4 rings Maximum how many codes can be formed to open the lock?
SHOW ANSWER
Correct Ans:9 x 9 x 9 x 9
Explanation:
9 x 9 x 9 x 9 = 94
12. How many words can be formed with or without meaning by taking all the letters from the word SMALL ?
SHOW ANSWER
Correct Ans:60
Explanation:
Number of letters in SMALL is 5, but there are 2 L?s Answer is 5! / 2! = 120 / 2 = 60
13. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
SHOW ANSWER
Correct Ans:720
Explanation:
The word 'OPTICAL' contains 7 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 5 letters can be arranged in 5! = 120 ways. The vowels (OIA) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6)
= 720.
14. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
SHOW ANSWER
Correct Ans:5040
Explanation:
LOGARITHMS' contains 10 different letters. Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= 10P4 = (10 x 9 x 8 x 7) = 5040.
15. In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
SHOW ANSWER
Correct Ans:36
Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5. Number of ways of arranging the vowels = 3P3 = 3! = 6. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3P3 = 3! = 6. Total number of ways = (6 x 6) = 36.
16. In how many ways can the letters of the word 'LEADER' be arranged ?
SHOW ANSWER
Correct Ans:360
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Therefore, Required number of ways = 6! / (1!)(2!)(1!)(1!)(1!) = 360
17. How many numbers greater than a million can be formed by using the digits 7, 4, 6 and 0 if 4 has to be used twice, 6 has to be used thrice and the rest only once?
SHOW ANSWER
Correct Ans:360
Explanation:
The given digits are 4, 4, 6, 6, 6, 7 and 0. Totally we have 7 digits. So numbers greater than a million can be formed by using all the digits. 4 occurs twice, 6 occurs thrice while 0 and 7 once, Therefore total number of arrangements = 7! / (3! x 2!) = 420 We have to avoid 0 in the starting place, Number of ways in which 0 can be in the first place is 6! / 2! X 3! = 60 Hence the total number of ways = 420 – 60 = 360
18. In how many ways a four digit even number can be formed by using the digits 4,5,9,8 exactly once.
SHOW ANSWER
Correct Ans:12
Explanation:
Given Four digit even number can be formed by using the digits 4,5,9,8 Since the number has to be a even digit number,the units digit has to be 4 or 8. First three places can be filled by remaining three digits. Hence it is totally 6 x 2 = 12 ways
Answer is 12
19. In how many ways a four digit even number can be formed by using the digits 2,3,5,8 exactly once?
SHOW ANSWER
Correct Ans:12
Explanation:
Solution is
Given
Four digit even number can be formed by using the digits 2,3,5,8
Since the number has to be an even number,the unit digit has to be 2 or 8.
First three places can be filled by remaining three digits.
Hence totally 12 ways
20. In a plane there are totally 8 points (no three points are collinear), how many lines can be drawn ?
SHOW ANSWER
Correct Ans:28
Explanation:
To draw a line we need 2 points. From the given 8 points , we need to choose 2 points Number of ways in which this can be done is 18C2= (8 x 7 ) / 2 = 28
Are you seeking for good platform for practicing Permutation Combination questions in online. This is the right place. The time you spent in Fresherslive will be the most beneficial one for you.
This page provides important questions on Permutation Combination along with correct answers and clear explanation, which will be very useful for various Interviews, Competitive examinations and Entrance tests. Here, Most of the Permutation Combination questions are framed with Latest concepts, so that you may get updated through these Permutation Combination Online tests. Permutation Combination Online Test questions are granted from basic level to complex level.
Permutation Combination questions are delivered with accurate answer. For solving each and every question, very lucid explanations are provided with diagrams wherever necessary.
Practice in advance of similar questions on Permutation Combination may improve your performance in the real Exams and Interview.
Time Management for answering the Permutation Combination questions quickly is foremost important for success in Competitive Exams and Placement Interviews.
Through Fresherslive Permutation Combination questions and answers, you can acquire all the essential idea to solve any difficult questions on Permutation Combination in short time and also in short cut method.
Winners are those who can use the simplest method for solving a question. So that they have enough time for solving all the questions in examination, correctly without any tense. Fresherslive provides most simplest methods to answer any tough questions. Practise through Fresherslive test series to ensure success in all competitive exams, entrance exams and placement tests.
Most of the job seekers finding it hard to clear Permutation Combination test or get stuck on any particular question, our Permutation Combination test sections will help you to success in Exams as well as Interviews. To acquire clear understanding of Permutation Combination, exercise these advanced Permutation Combination questions with answers.
You're Welcome to use the Fresherslive Online Test at any time you want. Start your beginning, of anything you want by using our sample Permutation Combination Online Test and create yourself a successful one. Fresherslive provides you a new opportunity to improve yourself. Take it and make use of it to the fullest. GOODLUCK for Your Bright Future.
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Permutation Combination Questions: Solved 27 Permutation Combination Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.
21. How many 4 digit even number can formed by using the digits 1,3,7 and 8 only once ?
SHOW ANSWER
Correct Ans:6
Explanation:
We can form 24 numbers with digits 1,3,7 and 8 using only once. To be an even number, the units digit has to even number, in this case only 8 should come in the unit's place . The remaining places, 3 positions can be filled in 3! , that is 6 ways
22. How many words can be formed with or without meaning by taking all the letters from the word TAKEN ?
SHOW ANSWER
Correct Ans:120
Explanation:
There are totally 4 letters in the word TAKEN T, A, K, E and N. No. Of Words = 5 x 4 x 3 x 2 x 1 = 120
23. In how many different ways can the letters of the word 'OFFICES' be arranged ?
SHOW ANSWER
Correct Ans:2520
Explanation:
The word OFFICES contains O,F, F I, C, E, S contains 7 letters, in which 2 are identical. No. Of words that can formed = 7 ! / 2 ! = 2520
SHOW ANSWER
Correct Ans:600
Explanation:
25! / 23! = 23! X 24 x 25 / ( 23!) = 24 x 25 = 600
25. In a party there were totally 20 people, each person shook his hands with the other person. How many hand shakes would have taken place ?
SHOW ANSWER
Correct Ans:190
Explanation:
There are 20 people. Every person has to shake hands with the other person. Which means we have to find the number of ways of choosing 2 people from the 20. The number of ways it ca happen = 20C2 = ( 20 x 19 ) / (1 x 2) = 190
26. A letter lock consists of 4 rings, each ring contains 9 non-zero digits. This lock can be opened by setting a 4 digit code with the proper combination of each of the 4 rings Maximum how many codes can be formed to open the lock?
SHOW ANSWER
Correct Ans:9 x 9 x 9 x 9
Explanation:
9 x 9 x 9 x 9 = 94
27. How many words can be formed with or without meaning by taking all the letters from the word SMALL ?
SHOW ANSWER
Correct Ans:60
Explanation:
Number of letters in SMALL is 5, but there are 2 L?s Answer is 5! / 2! = 120 / 2 = 60
Are you seeking for good platform for practicing Permutation Combination questions in online. This is the right place. The time you spent in Fresherslive will be the most beneficial one for you.
This page provides important questions on Permutation Combination along with correct answers and clear explanation, which will be very useful for various Interviews, Competitive examinations and Entrance tests. Here, Most of the Permutation Combination questions are framed with Latest concepts, so that you may get updated through these Permutation Combination Online tests. Permutation Combination Online Test questions are granted from basic level to complex level.
Permutation Combination questions are delivered with accurate answer. For solving each and every question, very lucid explanations are provided with diagrams wherever necessary.
Practice in advance of similar questions on Permutation Combination may improve your performance in the real Exams and Interview.
Time Management for answering the Permutation Combination questions quickly is foremost important for success in Competitive Exams and Placement Interviews.
Through Fresherslive Permutation Combination questions and answers, you can acquire all the essential idea to solve any difficult questions on Permutation Combination in short time and also in short cut method.
Winners are those who can use the simplest method for solving a question. So that they have enough time for solving all the questions in examination, correctly without any tense. Fresherslive provides most simplest methods to answer any tough questions. Practise through Fresherslive test series to ensure success in all competitive exams, entrance exams and placement tests.
Most of the job seekers finding it hard to clear Permutation Combination test or get stuck on any particular question, our Permutation Combination test sections will help you to success in Exams as well as Interviews. To acquire clear understanding of Permutation Combination, exercise these advanced Permutation Combination questions with answers.
You're Welcome to use the Fresherslive Online Test at any time you want. Start your beginning, of anything you want by using our sample Permutation Combination Online Test and create yourself a successful one. Fresherslive provides you a new opportunity to improve yourself. Take it and make use of it to the fullest. GOODLUCK for Your Bright Future.