Permutation Problems

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Permutation Problems

by MakeitHappen » Mon Feb 22, 2010 7:55 pm
Hey folks..am back with doubts abt how to solve more permutation problems..
Hope these questions also act as a good revision for those who are looking for more qns in this topic.

Qn1. How many even number of 4 digits can be formed with the figures 3,4,7,8 if repetitions are allowed?
4 and 8 are even numbers in the n no.of digits available(3,4,7,8). To form an even number, the number will have to end with 4 or 8. So, to select the last digit (the formula would be 2P2) and to select the remaining 3 digits the formula would be 3P3. The final answer IMO would be 2P2 * 3P3.
Actual answer: 128

Qn2. How many numbers greater than 4000 can be formed from the figures 3,5,7,8,9 ? (repetitions not allowed)?
Clueless on this one. But this is my understanding of the question: Numbs greater than 4000 mean the permutation would include number from 5378 to 98753. So will have to figure out how many are these?
Actual answer: 216

Qn3. The number of arrangements of 2n+2 different objects taken n at a time is to the number of arrangements of 2n different objects taken n at a time as 14:15. Find the value of n.
My understanding of this question:
(2n+2)Pn: (2n)Pn = 14:15
(2n+2)!/(2n+2-n) : (2n)!/(2n-n)!
(2n+2)!/(n+2)! : (2n)!/n! = 14 : 15
After this step, how to continue solving??
Actual answer:[spoiler] n = 3[/spoiler]

Qn 4. How many numbers of 7 digits can be formed from the digits 1,2,3,4,5,6,7 if there are not more than 2 digits between 1 and 2 ?
IMO:
the minimum num of digits that can be between 1 and 2 is only 1 digit.
Step 1:So, we would have to pick 1 digit out of the remaining 5 digits - 5P1
Step 2: Now that we have picked 1,2 and the digit inbetween 1 and 2, we need to find remaining number of 4 digits. For this, 4P4
Step 3: Treat 1, 2 and the digit in between 1 and 2 as one unit. SO we have remainig 4 units. So total, 5 units - 5P5
Step 4: Multiply: 5P1 * 4P4 * 5P5 = 14400
Actual answer: 15

Qn 5. In how many ways can 5 different mathematics books, 4 different physics books and 2 different chemistry books be arranged on a shelf if the books in each subject are to be together.
IMO:
Since each subject books will have to be together treat as 3 units instead of 5+4+2=11 units. So, 3P3
Now among the 5 math book there are difft wayts of arrangement : 5P5. Same goes for physics and chemisttry books.
Therefore IMO answer shd be 3P3 * 5P5 * 4P4 * 2P2 = 28800
Actual answer: 34560

Qn 6. Find the number of arrangements of the letter in the word PENCILS if (i) E precedes I, (ii) there are three letters between E and I.
Actual answer : (i) 2520 (ii) 120

Qn 7. In how many ways can the letters of the word PRINCIPLE be arranged? In what proportion of these arrangements do the letters P come together ?
I thought this was a fairly simple question! There are 9 letters in the word PRINCIPLE. So IMO 9P9 is the answer.
Actual answer: 90720

Qn 8. In how many ways can the letters in PRECISION be arranged? In how many of these arrangements do the vowels occupy even places ?
Actual answer: 181440

Qn 9. How many arrangements can be made by the letters of word DEFINITION (i) if the letters I do not occupy the first or last place, (ii) if the letters I are together?
Actual answer: (i) 141120 (ii) 20160

Qn 10. How many different arrangements of the letters in TOMATO are there, if the letters O are to be separated?
Actual answer: 120

Qn 11. A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?
2 particular people must sit in the back seat and 1 particular person is the driver. therefore, only remaining 2 pple are left to be seated in the back seat and so, out of 7 choices, we only have 4 choices. So, 4P2.
1 particular person is the driver, so there are 3 pple that we need to select from the remaining 3 pple. So, 3P3
Multiple 4P2 * 3P3
Actual answer: 288
Source: — Problem Solving |

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by thephoenix » Mon Feb 22, 2010 8:47 pm
MakeitHappen wrote:Hey folks..am back with doubts abt how to solve more permutation problems..
Hope these questions also act as a good revision for those who are looking for more qns in this topic.

Qn1. How many even number of 4 digits can be formed with the figures 3,4,7,8 if repetitions are allowed?
4 and 8 are even numbers in the n no.of digits available(3,4,7,8). To form an even number, the number will have to end with 4 or 8. So, to select the last digit (the formula would be 2P2) and to select the remaining 3 digits the formula would be 3P3. The final answer IMO would be 2P2 * 3P3.
Actual answer: 128
repetition is allowed and the no. has to be even
for it to be evn the unit dig should be even i.e either 4 or 8 so 2 ways

rest 3 place has 4 choices equally available(bcoz repetition is allowed)=4*4*4=64

so total ways=2*64=128

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by thephoenix » Mon Feb 22, 2010 9:03 pm
MakeitHappen wrote:Hey folks..am back with doubts abt how to solve more permutation problems..
Hope these questions also act as a good revision for those who are looking for more qns in this topic.


Qn2. How many numbers greater than 4000 can be formed from the figures 3,5,7,8,9 ? (repetitions not allowed)?
Clueless on this one. But this is my understanding of the question: Numbs greater than 4000 mean the permutation would include number from 5378 to 98753. So will have to figure out how many are these?
Actual answer: 216
4 digit no. not starting with 3 and 5 digits no.
there are two cases a
a four digit no. starting with 3 will always be < 4000 so we have only 4 choice for 1000nth place so #of ways=4C1=4
now out of the remaining 4 digits the # of arrangements can be =4P3=4!=24
total 4digits no.=4*24=96

a five digit no. can be formed in 5P3 ways=5!=120

total=120+96=216

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by papgust » Mon Feb 22, 2010 9:38 pm
Makeithappen,

Please post a single question in a thread!

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by harsh.champ » Mon Feb 22, 2010 10:49 pm
MakeitHappen wrote:Hey folks..am back with doubts abt how to solve more permutation problems..
Hope these questions also act as a good revision for those who are looking for more qns in this topic.

Qn3. The number of arrangements of 2n+2 different objects taken n at a time is to the number of arrangements of 2n different objects taken n at a time as 14:15. Find the value of n.
My understanding of this question:
(2n+2)Pn: (2n)Pn = 14:15
(2n+2)!/(2n+2-n) : (2n)!/(2n-n)!
(2n+2)!/(n+2)! : (2n)!/n! = 14 : 15
After this step, how to continue solving??
Actual answer:[spoiler] n = 3[/spoiler]
(2n+2)!/(n+2)! : (2n)!/n! = 14 : 15
=>[(2n+2)! x n!]/[(n+2)! x 2n!] = 14:15
=>[(2n+2)(2n+1)]/[(n+2)(n+1)] =14:15
=>[2(n+1)(2n+1)]/[(n+2)(n+1)] = (2 x 7):15
=>(2n+1)/(n+2) = 7:15

Solving u get n=3
I hope it is clear now.
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by harsh.champ » Mon Feb 22, 2010 10:58 pm
MakeitHappen wrote:
Qn 4. How many numbers of 7 digits can be formed from the digits 1,2,3,4,5,6,7 if there are not more than 2 digits between 1 and 2 ?
IMO:
the minimum num of digits that can be between 1 and 2 is only 1 digit. -How can you say this??
Here' an example when 2 digits are there1342567,1452367

Step 1:So, we would have to pick 1 digit out of the remaining 5 digits - 5P1
Step 2: Now that we have picked 1,2 and the digit inbetween 1 and 2, we need to find remaining number of 4 digits. For this, 4P4
Step 3: Treat 1, 2 and the digit in between 1 and 2 as one unit. SO we have remainig 4 units. So total, 5 units - 5P5
Step 4: Multiply: 5P1 * 4P4 * 5P5 = 14400
Actual answer: 15
We have to consider 2 cases:-
Case 1:- When no digits are there b/w 1 and 2 :- 6! x 2!
Case 2:- When 1 digit is there :- 5C1 x 5! x 2!
Case 3:- When 2 digits are there

Kindly check the question.
Either something is missing in it or the option choices are wrong.
Whats the source?
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by harsh.champ » Mon Feb 22, 2010 11:01 pm
MakeitHappen wrote:
Qn 5. In how many ways can 5 different mathematics books, 4 different physics books and 2 different chemistry books be arranged on a shelf if the books in each subject are to be together.
IMO:
Since each subject books will have to be together treat as 3 units instead of 5+4+2=11 units. So, 3P3
Now among the 5 math book there are difft wayts of arrangement : 5P5. Same goes for physics and chemisttry books.
Therefore IMO answer shd be 3P3 * 5P5 * 4P4 * 2P2 = 28800
Actual answer: 34560
Its a calculation mistake.
3! x 5! x 4! x 2! = 34560
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by harsh.champ » Mon Feb 22, 2010 11:11 pm
MakeitHappen wrote:
Qn 6. Find the number of arrangements of the letter in the word PENCILS if (i) E precedes I, (ii) there are three letters between E and I.
Actual answer : (i) 2520 (ii) 120
1)E precedes I :-
a)E is in the 1st posn- I can take any of the 6 succeeding posns. - 6C1
Rest of the letters in 5! ways.
Total = 6C1 x 5! ways

Like this for subsequent cases you will have :-
b)5C1 x 5!
c)4C1 x 5!
d)3C1 x 5!
e)2C1 x 5!
f)1C1 x 5!

Adding all the cases,we get 21 x 5! =2520 ans.


2)There are 3 lettes b/w E and I:-
Those 3 letters can be selected in 5C3 ways.
Let E,I,those letters be a separate entity.
In that entity,those 3 letters can be interchanged in 3! ways.
E and I can be interchanged in 2! ways
Entity + rest of the 2 letters can be interchanged in 3! ways
Hence, ans. = 5C3 x 3! x 2! x 3!=720

I think it should be 720 not 120 can you check plz
120 = 5! I dont know how you got 5! as the answer over here??
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by harsh.champ » Tue Feb 23, 2010 2:19 am
MakeitHappen wrote:
Qn 7. In how many ways can the letters of the word PRINCIPLE be arranged? In what proportion of these arrangements do the letters P come together ?
I thought this was a fairly simple question! There are 9 letters in the word PRINCIPLE. So IMO 9P9 is the answer.
Actual answer: 90720
Hey makeithappen,
You forgot that there are 2 I's and 2 P's in the word PRINCIPLE.
Hence,answer would be 9!/(2! x 2!) = 362880/4 = 90720

Hope it would be clear now.
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by arzanr » Tue Feb 23, 2010 5:27 pm
harsh.champ wrote:
MakeitHappen wrote:Hey folks..am back with doubts abt how to solve more permutation problems..
Hope these questions also act as a good revision for those who are looking for more qns in this topic.

Qn3. The number of arrangements of 2n+2 different objects taken n at a time is to the number of arrangements of 2n different objects taken n at a time as 14:15. Find the value of n.
My understanding of this question:
(2n+2)Pn: (2n)Pn = 14:15
(2n+2)!/(2n+2-n) : (2n)!/(2n-n)!
(2n+2)!/(n+2)! : (2n)!/n! = 14 : 15
After this step, how to continue solving??
Actual answer:[spoiler] n = 3[/spoiler]
(2n+2)!/(n+2)! : (2n)!/n! = 14 : 15
=>[(2n+2)! x n!]/[(n+2)! x 2n!] = 14:15
=>[(2n+2)(2n+1)]/[(n+2)(n+1)] =14:15
=>[2(n+1)(2n+1)]/[(n+2)(n+1)] = (2 x 7):15
=>(2n+1)/(n+2) = 7:15

Solving u get n=3
I hope it is clear now.

How do you get from =>[(2n+2)! x n!]/[(n+2)! x 2n!] = 14:15 to =>[(2n+2)(2n+1)]/[(n+2)(n+1)] =14:15 ??

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by arzanr » Tue Feb 23, 2010 5:47 pm
harsh.champ wrote:
MakeitHappen wrote:
Qn 6. Find the number of arrangements of the letter in the word PENCILS if (i) E precedes I, (ii) there are three letters between E and I.
Actual answer : (i) 2520 (ii) 120
1)E precedes I :-
a)E is in the 1st posn- I can take any of the 6 succeeding posns. - 6C1
Rest of the letters in 5! ways.
Total = 6C1 x 5! ways

Like this for subsequent cases you will have :-
b)5C1 x 5!
c)4C1 x 5!
d)3C1 x 5!
e)2C1 x 5!
f)1C1 x 5!

Adding all the cases,we get 21 x 5! =2520 ans.


2)There are 3 lettes b/w E and I:-
Those 3 letters can be selected in 5C3 ways.
Let E,I,those letters be a separate entity.
In that entity,those 3 letters can be interchanged in 3! ways.
E and I can be interchanged in 2! ways
Entity + rest of the 2 letters can be interchanged in 3! ways
Hence, ans. = 5C3 x 3! x 2! x 3!=720

I think it should be 720 not 120 can you check plz
120 = 5! I dont know how you got 5! as the answer over here??

I got 720 as well for ii.

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by gmat_basher » Fri Feb 26, 2010 1:58 pm
arzanr wrote:
harsh.champ wrote:
MakeitHappen wrote:
Qn 6. Find the number of arrangements of the letter in the word PENCILS if (i) E precedes I, (ii) there are three letters between E and I.
Actual answer : (i) 2520 (ii) 120
1)E precedes I :-
a)E is in the 1st posn- I can take any of the 6 succeeding posns. - 6C1
Rest of the letters in 5! ways.
Total = 6C1 x 5! ways

Like this for subsequent cases you will have :-
b)5C1 x 5!
c)4C1 x 5!
d)3C1 x 5!
e)2C1 x 5!
f)1C1 x 5!

Adding all the cases,we get 21 x 5! =2520 ans.


2)There are 3 lettes b/w E and I:-
Those 3 letters can be selected in 5C3 ways.
Let E,I,those letters be a separate entity.
In that entity,those 3 letters can be interchanged in 3! ways.
E and I can be interchanged in 2! ways
Entity + rest of the 2 letters can be interchanged in 3! ways
Hence, ans. = 5C3 x 3! x 2! x 3!=720

I think it should be 720 not 120 can you check plz
120 = 5! I dont know how you got 5! as the answer over here??

I got 720 as well for ii.
I think the answer is 120, you are also counting letters b/w I and E instead of only E and I.

E and I can occur only in postion 1 ,5 2,6 and 3,7 ( so 3 ways )

3 * 5c3 * 2 = 120

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by gmat_basher » Fri Feb 26, 2010 2:13 pm
MakeitHappen wrote: Qn 11. A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?
2 particular people must sit in the back seat and 1 particular person is the driver. therefore, only remaining 2 pple are left to be seated in the back seat and so, out of 7 choices, we only have 4 choices. So, 4P2.
1 particular person is the driver, so there are 3 pple that we need to select from the remaining 3 pple. So, 3P3
Multiple 4P2 * 3P3
Actual answer: 288
Assume the 7 people are

A B C D E F G and H

Let D be driver and G,H people who should sit in the back

With D in driver seat, and GH in back, there are 4C2 choice for front seat from A C E F , and 4C3 Choice for back seat ,
4c2 * 4c3 = 288

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by ncc800 » Sun Apr 22, 2012 6:50 pm
(2n+2)!/(n+2)! : (2n)!/n! = 14 : 15
=>[(2n+2)! x n!]/[(n+2)! x 2n!] = 14:15
=>[(2n+2)(2n+1)]/[(n+2)(n+1)] =14:15
=>[2(n+1)(2n+1)]/[(n+2)(n+1)] = (2 x 7):15
=>(2n+1)/(n+2) = 7:15

Solving u get n=3
I hope it is clear now.

What am I missing here? when I solve (2n+1)/(n+2) = 7:15 I do not get n=3