Permutations and Combinations #1

[mishum2000]

1) A man has seven relatives, 4 women and 3 men. His wife also has 7 relatives, 3 men and 4 women. In how many ways can they have a dinner party of 3 men and 3 women so that there are 3 of the man’s relatives and 3 of his wife’s relatives?

2) The number of successful attempts that can be made by a thief to open a number lock having three rings of which each ring contains 9 numbers is ?

3) The no. of unsuccessful attempts that can be made by a thief to open a number lock having four rings In which each ring contains 4 numbers is ?

4) The number of even numbers between 1000 and 10000 can be formed by the numbers 1 2 3 4 5 6 7 8 9 is ?

5) The number of numbers not divisible by 5 that can be made from the digits 0 2 4 5

6) The number of arrangements that can be made from the word MISSISSIPI so that all the S’s come together and the I’s don’t come together is ?

[jose]

1) A man has seven relatives, 4 women and 3 men. His wife also has 7 relatives, 3 men and 4 women. In how many ways can they have a dinner party of 3 men and 3 women so that there are 3 of the man’s relatives and 3 of his wife’s relatives?

2) The number of successful attempts that can be made by a thief to open a number lock having three rings of which each ring contains 9 numbers is ?

3) The no. of unsuccessful attempts that can be made by a thief to open a number lock having four rings In which each ring contains 4 numbers is ?

4) The number of even numbers between 1000 and 10000 can be formed by the numbers 1 2 3 4 5 6 7 8 9 is ?

5) The number of numbers not divisible by 5 that can be made from the digits 0 2 4 5

6) The number of arrangements that can be made from the word MISSISSIPI so that all the S’s come together and the I’s don’t come together is ?

PROBLEM 1

Let x: number of men who will attend the party
y: number of women who will attend the party

man (x1,y1)-------wife (x2,y2)
x1+y1=3 and x2+y2=3

case A: x1,y1 (3,0)-----x2,y2 (0,3)

a=3c3*4c3=4

case B: x1,y1 (2,1)-----x2,y2 (1,2)

b=3c2*4c1*3c1*4c2=216

case C: x1,y1 (1,2)-----x2,y2 (2,1)

c=3c1*4c2*3c2*4c1=216

case D: x1,y1 (0,3)-----x2,y2 (3,0)

d=4c3*3c3=4

answer= a+b+c+d = 440


PROBLEM 4

Numbers with these forms is what we are looking for:
xxx2, xxx4, xxx6 and xxx8

4*8p3 = 1344


PROBLEM 5

20 (?)

PROBLEM 6

4!*7!-4!^3 = 107136

[Top_Gun]

1) A man has seven relatives, 4 women and 3 men. His wife also has 7 relatives, 3 men and 4 women. In how many ways can they have a dinner party of 3 men and 3 women so that there are 3 of the man’s relatives and 3 of his wife’s relatives?


Answer: There can be 4 cases:

Man's Relatives Wife's Relative

case 1: 3 woman 3 men

case 2: 2 women, 1 man 1 woman, 2 men

case 3: 1 woman, 2 man 2 women, 1 man

case 4: 3 men 3 women

So, the number of ways:

For case 1 and 4 each : 3c3 * 4c3 = 4

For case 2 and 3 each: 4c2* 3c1 * 4c1* 3c2 = 216

So, total number of ways = 4 + 216 + 216 + 4 = 440


2) The number of successful attempts that can be made by a thief to open a number lock having three rings of which each ring contains 9 numbers is ?

Answer: 1 *1 *1 = 1 way.

3) The no. of unsuccessful attempts that can be made by a thief to open a number lock having four rings In which each ring contains 4 numbers is ?


Answer: The total number of combinations is: 4 * 4 * 4 * 4
= 256
Here, one attempt will be successful.

So, the number of unsuccessful attempts is : 256 - 1
= 255
4) The number of even numbers between 1000 and 10000 can be formed by the numbers 1 2 3 4 5 6 7 8 9 is ?


Answer: This is virtually all 4 digit numbers except for the lack of 0.

anyway, the number = 9 * 9 * 9 * 4 = 2916


5) The number of numbers not divisible by 5 that can be made from the digits 0 2 4 5

Answer: Here, repetition of numbers should not be allowed or else there would be infinite of such numbers.

1 digit number: 2
2 digit number : 2 * 2 = 4
3 digit number : 2 * 3 * 2( including 0 in the first digit ) - 2 * 2( only 0 in the first place )
= 8

4 digit number: 1 * 2 * 3 * 2 - 1 * 1 * 2 * 2 = 8

So, total such numbers : 22

6) The number of arrangements that can be made from the word MISSISSIPI so that all the S’s come together and the I’s don’t come together is ?

Answer: The number of arrangements that S come together = 7!/ 4!
= 210

among these arrangements I comes together in = 3! arrangements
= 6 arrangements.

So, total such arrangements = 210 - 6
= 204

[Top_Gun]

6) The number of arrangements that can be made from the word MISSISSIPI so that all the S’s come together and the I’s don’t come together is ?

Answer: The number of arrangements that S come together = 7!/ 4!
= 210

among these arrangements I comes together in = 4! arrangements
= 24 arrangements.

So, total such arrangements = 210 - 24
= 186

[denizen06]

Hi Top Gun,
Great job with workings 1.
But I am little bi confused with no. 5.
Shouldn't we use all the digits together?
In that case the ans should be 8
Please help me to find the falults of my reasonoing

5) The number of numbers not divisible by 5 that can be made from the digits 0 2 4 5

Answer: Here, repetition of numbers should not be allowed or else there would be infinite of such numbers.

1 digit number: 2
2 digit number : 2 * 2 = 4
3 digit number : 2 * 3 * 2( including 0 in the first digit ) - 2 * 2( only 0 in the first place )
= 8

4 digit number: 1 * 2 * 3 * 2 - 1 * 1 * 2 * 2 = 8

So, total such numbers : 22

[wydiwyg]

6) The number of arrangements that can be made from the word MISSISSIPI so that all the S’s come together and the I’s don’t come together is ?

Answer: The number of arrangements that S come together = 7!/ 4!
= 210

among these arrangements I comes together in = 4! arrangements
= 24 arrangements.

So, total such arrangements = 210 - 24
= 186

shold not the all S's come together be this way?

MISSISSIPI
SSSS
SSSS
SSSS
SSSS
SSSS
SSSS
SSSS
SSSS
Total of 8 ways. There is no other ways that all S's can come together.

How can it be that
The number of arrangements that S come together = 7!/ 4!
= 210


I cant get the second one that all I's cant come together.
among these arrangements I comes together in = 4! arrangements
= 24 arrangements.
Top gun please answer

[Top_Gun]

If we are to find all the permutations where S has to come together, we can simply assume all the S's( 4 S's ) to be just one character. Then all S would come together.

So, now there r 7 characters( M, P, 4 I's and a single S ).
So, number of permutations = 7!/4!

Again, the question asks us to find out the numbers where none of the I's would come together.

So, we've to just find the number of such combinations where I's come together.

Now, likewise, we'ld think that all the I's are also a single I.
So, the number of characters is: 4( M, P, only a solitary I and a solitary S ) and all of them r distinct characters.

So, number of such characters : 4!

So, answer is = 7!/4! - 4!

shold not the all S's come together be this way?

MISSISSIPI
SSSS
SSSS
SSSS
SSSS
SSSS
SSSS
SSSS
SSSS
Total of 8 ways. There is no other ways that all S's can come together.

How can it be that
The number of arrangements that S come together = 7!/ 4!
= 210


I cant get the second one that all I's cant come together.
among these arrangements I comes together in = 4! arrangements
= 24 arrangements.
Top gun please answer

[Top_Gun]

Hi Top Gun,
Great job with workings 1.
But I am little bi confused with no. 5.
Shouldn't we use all the digits together?
In that case the ans should be 8
Please help me to find the falults of my reasonoing

Well, in that case, i expect some clear cut indications in the question. As no such indications were given in the question, i had to solve the question that way. thy answer may be right, but it all depends upon what the question makers thought while making the question, not what is meant by the question.

[denizen06]

Yah. Right you are. Who knows what is driving the questionmaker's brain?
In that case we need to be lucky. Don't we ?

Well, in that case, i expect some clear cut indications in the question. As no such indications were given in the question, i had to solve the question that way. thy answer may be right, but it all depends upon what the question makers thought while making the question, not what is meant by the question.

[Top_Gun]

Well, we need to be lucky, but u don't.

Well, u seem like a math wizard. That's why they have left enough leeway in the scoring system to ensure perfect score for u guys. We should rather be worried, we'ld fail to answer some maths cause we can't and also some cause the question might not be a well articulated one.