Integral Solutions

Probability Level 5

$abcde=1050$

and a, b, c, d, e are positive integers. How many ordered tuples $(a, b, c, d, e)$ exist?

Extra Credit: How many unordered sets ${a, b, c, d, e}$ exist?

Extra Credit: How many ordered tuples $(a, b, c, d, e)$ exist such that a≠b≠c≠d≠e?

Unrelated Credit: Solve the Aunty's Teacups problem

The answer is 1875.

1 solution

Agnishom Chattopadhyay
Sep 3, 2014

You have a box [2,3,5,5,7]

You have initialised a=b=c=d=e=1

You have to take away a number from the box and multiply it with one of a, b, c or d and repeat until the box is exhausted.

In how many ways can you do this?

The numbers 2,3,7 can be multiplied with any of a,b,c,d,e. So there are 5^3 ways of doing it.

Either you multiply 5^2 to one of the numbers which can be done in 5 ways or you throw the two 5's at two two different variables which can be done in 10 ways.

So you can do it in a total of 5^3(10+5) ways.

The easier way to express your ideas, is that we want to find non-negative integer solutions to:

$\begin{array} {l l} a_2 + b_2 + c_2 + d_2 + d_2 & = 1\\ a_3 + b_3 + c_3 + d_3 + d_3 & = 1 \\ a_5 + b_5 + c_5 + d_5 + d_5 & = 2 \\ a_7 + b_7 + c_7 + d_7 + d_7 & = 1 \\ \end{array}$

where $a_n$ denotes the prime power exponent of $n$ in $a$ .

Hence, there are ${ 5 \choose 1 } ^3 \times { 6 \choose 2 }$ ways.

Calvin Lin Staff - 6 years, 9 months ago

Exactly How I did it.

Ronak Agarwal - 6 years, 9 months ago

Me too Exactly same.

Gautam Sharma - 6 years, 6 months ago

what if we were asked to find number of unordered tuples???

Vighnesh Raut - 6 years, 4 months ago

I Reason it in this way : $\displaystyle{abcde=2\times 3\times { 5 }^{ 2 }\times 7}$
Now This situation is analogous to, There are

: 1 book type-1 (say 2)

: 1 book type-2 (say 3)

: 2 book type-3 (say 5)

: 1 book type-4 (say 7)

"Here Book of Same type are identical "

and there are 5 beggar's , and we have to distribute these books to them .

Book-1 can be distributed to 5 beggers in $(\begin{matrix} 5 \\ 4 \end{matrix})$ ways !

Book-2 . . . . . . . . . . . . . . $\displaystyle{(\begin{matrix} 5 \\ 4 \end{matrix})}$ ways !

Book-3 ............................ $\displaystyle{(\begin{matrix} 6 \\ 4 \end{matrix})}$ ways!

Book-4........................... $\displaystyle{(\begin{matrix} 5 \\ 4 \end{matrix})}$ ways !

So Total no of ways = $\displaystyle{{ (\begin{matrix} 5 \\ 4 \end{matrix}) }^{ 3 }\times (\begin{matrix} 6 \\ 4 \end{matrix})}$

Deepanshu Gupta - 6 years, 4 months ago

Integral Solutions

The answer is 1875.

1 solution

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