Question

the total number of positive integral solutions of the equation x1.x2.x3.x4=10!

Answer 1

yes

Answer 2

x1.x2.x3.x4=3628800

Answer 3

@ganeshie8

Answer 4

@zepdrix @Compassionate @iambatman

Answer 5

@arabpride

Answer 6

no one knows the answer i think

Answer 7

@Jhannybean

Answer 8

Nope -No idea tbh~ cx

Answer 9

Is the way @perl solved it wrong?..

Answer 10

yes it is wrong

Answer 11

As a start : find the prime factorization of 10!

Answer 12

then after that

Answer 13

It will be easy if I walk you through

Answer 14

whats the prime factorization of 10! ?

Answer 15

thats not prime factorization

Answer 16

its factorial

Answer 17

\[\rm 10! = 2^8\times 3^4\times 5^2\times 7^1\]

Answer 18

yes then

Answer 19

Consider a divisor of \(10!\), It will be of form :
\[\rm 2^a\times 3^b\times 5^c\times 7^d\]

where \(\rm 0\le a\le 8, ~0\le b\le 4, ~0\le c\le 3, ~0\le d\le 1\)

yes ?

Answer 20

ok.then

Answer 21

o<=c<=2

Answer 22

Oh rihgt, thanks for correcting @Princer_Jones :)

Answer 23

ok.then

Answer 24

we're not done yet,
we need to count number of solutions for (a,b,c,d) based on these divisors

Answer 25

@ganeshie8 I am learning...:)

Answer 26

its good princer

Answer 27

ganeshie i m waiting

Answer 28

Patience, young grasshopper.

Answer 29

ok miss jhanny

Answer 30

ok.every question i post here is tricky bt i hope u will solve it

Answer 31

jhanny u cn also try

Answer 32

let me tag few others @nerdguy2535 @ChristopherToni @dan815

Answer 33

@rishavraj

Answer 34

Say \[\rm x_i = 2^{a_i}\times 3^{b_i}\times 5^{c_i}\times 7^{d_i} \]

and look at your equation
\[\large \rm x_1\cdot x_2\cdot x_3\cdot x_4 = 2^8\cdot 3^4\cdot 5^2\cdot 7^1\]
Clearly each exponent of a prime, \(p\) in product \(\rm x_1\cdot x_2\cdot x_3\cdot x_4\) must equal to the exponent of \(p\) in \(10!\) :

\[a_1 + a_2 + a_3 + a_4 = 8 \\
b_1 + b_2 + b_3 + b_4 = 4 \\
c_1+c_2+c_3+c_4 = 2 \\
d_1+d_2+d_3+d_4 = 1
\]

find the number of nonnegative solutions of each of above equations and multiply them.

Answer 35

using stars and bars you get \[\large \rm \binom{11}{3}\cdot \binom{7}{3}\cdot \binom{5}{3}\cdot \binom{4}{3}\]

Answer 36

http://www.wolframalpha.com/input/?i=+%5Cbinom%7B11%7D%7B3%7D%5Ccdot+%5Cbinom%7B7%7D%7B3%7D%5Ccdot+%5Cbinom%7B5%7D%7B3%7D%5Ccdot++%5Cbinom%7B4%7D%7B3%7D

Answer 37

umm prolly too late, but just to confirm,
you take 10! take all its prime factors, and see how many ways to get unique arrangements of that, then take stars and bars for each arrangement - then subtract the possible overlaps
just wondering if there is a way in this too, i feel like the overlaps get ugly,

i do like this solution though its nice
say u have
2,2,2,3,3,3,5,5,7,7,7,7

_ _ _ _ . and you throw the number 2s around in there. For each arrangement of 2s in the bars there a different arrangemnt of 3s and 5s and 7s that will all multiply to 10!
so the answer is (number of ways to split up the 3s between 4 bars) * (#of2s btwn 4 bars) * (#7btwen4bars...* and so on)

Answer 38

@ganeshie8 can you explain the method, or post me any link of the theory section concerning this kind of problems.?? Thanks.:)

Answer 39

the 2nd part of my post is what ganeshie did

Answer 40

the process of splitting the number of 2s between 4 places is, the stars and bars method, like u have 8 2s, so
p1+p2+p3+p4 = 8, where p1 to p4 cannot have a negative integer

Answer 41

2 2 2 2 2 2 2 2 <------------8 2s how to split into 4 place
2 2 2 | 2|2 2 | 2 2 <--------introduce these 3 bars, or 3 extra places
so 11 Choose 3, u think of it as placing the 3 bars around in any of the 11 places now as there are 11 objects with the 2s and the bars in total now

Answer 42

so thats the total number of ways to split up the 2s, and for each split of 2 there are a differnt number of splits of 3s 5s and 7s that work, so u are multipling those arrangements together with this to get the total number of possible ways

Answer 43

@Princer_Jones i like this explanation here about stars and bars
https://brilliant.org/discussions/thread/stars-and-bars/

Answer 44

lets do an exaple maybe

Answer 45

find the number of nonnegative integral solutions to the equation :
\[\large \rm x_1 + x_2 + x_3 = 4\]

Answer 46

OK...

Answer 47

we translate this problem by creating a bijection to a similar "stars and bars" problem

Answer 48

there were sequences of length 13 in that problem

Answer 49

yes lets work the same problem maybe

\[\large \rm a+b+c+d = 10\]

Answer 50

yeah

Answer 51

finding number of nonnegative integral solutions to above equation is equivalent to finding number of ways of arranging bars :
\[\large \rm ***|***|***|*\]

Answer 52

number of stars before first bar gives you the value of \(a\),

number of stars before first bar and second bar gives you the value of \(b\) etc..

Answer 53

In that string of stars and bars of length 13, you can choose 3 positions for bars in \(\large \binom{13}{3}\) ways.

Answer 54

***|***|***|* are we to put the bars in this fashion only?

Answer 55

that gives one solution : (a,b,c,d) = (3,3,3,1)

Answer 56

\[\large \rm **|****|*|*** ~~~~\sim~~~ (2,4,1,3) \]

Answer 57

\[\large \rm **|****|****| ~~~~\sim~~~ (2,4,4,0) \]

Answer 58

oh...

Answer 59

there is an one-to-one association between "stars and bars" problem and the equation we are solving

Answer 60

hmm correct..

Answer 61

there are 10 stars and 3 bars, so the length of string is : 10+3 = 13

Answer 62

you can choose any 3 locations from those 13 locations for placing ur bars

Answer 63

hmmm next?

Answer 64

the total number of different strings with 10 stars and 3 bars is :
\[\large \rm \binom{13}{3}\]

we're done.

Answer 65

oh... this kind of problem falls under which branch of mathematics?

Answer 66

number theory

Answer 67

under life

Answer 68

basically what we're doing is translating an "unknown problem"(# of solutions to equation) to a known problem("stars and bars") by creating a bijection between the two

Answer 69

oh i see.. :)....

Answer 70

lets do another example if u have time

Answer 71

yeah i hv time..

Answer 72

\[\large \rm x_1 + x_2 + x_3 = 4\]

Answer 73

find number of nonnegative integral solutions to above eqn

Answer 74

now that we have "stars and bars" trick, we just translate the problem and follow our noses

Answer 75

how many stars ?
how many bars ?

Answer 76

find number of nonnegative integer* solutions to above eqn

Answer 77

4

Answer 78

4stars and 2 bars i gues..

Answer 79

yes because you want the left hand side things to add up to 4

you need bars to divide the stars

Answer 80

yes!
\[\large \rm *|**|*\]

Answer 81

hmm It should be 6C2 ???

Answer 82

Correct !

Answer 83

you can choose 2 positions for our bars from the available 6 positions in 6C2 ways

Answer 84

first one 13C2 and second one 9C2/@dan815

Answer 85

no the point is the bar is cutting both sets of stars at hte same time

Answer 86

@dan815 i cant see the picture anymore.. and couldnot get your point..

Answer 87

-.-

Answer 88

@ganeshie8 so using this stars and bars method we find the number of possible combinations of the 4 linear equations you establshed before, and they multiply them to get the final solution?

Answer 89

me neither

Answer 90

cuz im making stuff up

Answer 91

yes, you will feel "stars and bars" pretty mechanical once you get use to it...

Answer 92

\[a_1 + a_2 + a_3 + a_4 = 8 \\
b_1 + b_2 + b_3 + b_4 = 4 \\
c_1+c_2+c_3+c_4 = 2 \\
d_1+d_2+d_3+d_4 = 1\]

Answer 93

I bet you can work all above equations without much thinking

Answer 94

Yeah obviously.:)

Answer 95

(8+3 C 3)=11C3, (4+3C3)=7C3,...

Answer 96

\[\large \rm \binom{11}{3}\cdot \binom{7}{3}\cdot \binom{5}{3}\cdot \binom{4}{3}\]