Counting, nPr and nCr problems

ahmedcrow

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I'll make this thread for problems of counting principles, permutations, and combinations if it's allowed, because they're one unit in the book I'm studying, I read some about LaTex that can help me to form some problems, I'll translate the Arabic mathematical notations of problems I can't solve.

#1: if \(\displaystyle ^nP_4 = 405n\) so n = ?

The answer is n=10 but I don't how, they don't explain the way in the book.
 
I'll make this thread for problems of counting principles, permutations, and combinations if it's allowed, because they're one unit in the book I'm studying, I read some about LaTex that can help me to form some problems, I'll translate the Arabic mathematical notations of problems I can't solve.

#1: if \(\displaystyle ^nP_4 = 405n\) so n = ?

The answer is n=10 but I don't how, they don't explain the way in the book.

\(\displaystyle ^{n}P_{4} = \dfrac{n!}{(n-4)!} = n(n-1)(n-2)(n-3) = 405n\)


\(\displaystyle (n-1)(n-2)(n-3) = 405\)


Are you SURE n = 10?
Is it really "="? Maybe ">"?
 
I'll make this thread for problems of counting principles, permutations, and combinations if it's allowed, because they're one unit in the book I'm studying, I read some about LaTex that can help me to form some problems, I'll translate the Arabic mathematical notations of problems I can't solve.

#1: if \(\displaystyle ^nP_4 = 405n\) so n = ?

The answer is n=10 but I don't how, they don't explain the way in the book.

You can check the answer and it works. As you have been shown, you would have to solve a cubic equation. It is much easier just to use trial and error (guess and check), since n can't be very big, and must be an integer.
 
\(\displaystyle ^{n}P_{4} = \dfrac{n!}{(n-4)!} = n(n-1)(n-2)(n-3) = 405n\)


\(\displaystyle (n-1)(n-2)(n-3) = 405\)


Are you SURE n = 10?
Is it really "="? Maybe ">"?

Yes n=10 but I'm sorry \(\displaystyle (n-1)(n-2)(n-3) = 504\) instead of \(\displaystyle (n-1)(n-2)(n-3) = 405\) it was a fault in writing.

You can check the answer and it works. As you have been shown, you would have to solve a cubic equation. It is much easier just to use trial and error (guess and check), since n can't be very big, and must be an integer.

I think there's no cubic equations in our secondary schools, we need to solve the problem without it, I think I found the way.

This's my suggestion:

\(\displaystyle ^nP_r = n(n-1)(n-2)...(n-r+1)\)

and

\(\displaystyle (n-1)(n-2)(n-3) = ^{(n-1)}P_r = 504\)
\(\displaystyle (n-1)-r+1 = n-3 \) the last factor
\(\displaystyle n-n-r=-3\)
\(\displaystyle r=3 \)

so \(\displaystyle ^{(n-1)}P_r = ^{(n-1)}P_3 = 504\)
\(\displaystyle ^{(n-1)}P_3 = ^9P_3 = 504\)

so \(\displaystyle n-1 = 9\)
\(\displaystyle n = 10\)
 
Yes n=10 but I'm sorry \(\displaystyle (n-1)(n-2)(n-3) = 504\) instead of \(\displaystyle (n-1)(n-2)(n-3) = 405\) it was a fault in writing.



I think there's no cubic equations in our secondary schools, we need to solve the problem without it, I think I found the way.

This's my suggestion:

\(\displaystyle ^nP_r = n(n-1)(n-2)...(n-r+1)\)

and

\(\displaystyle (n-1)(n-2)(n-3) = ^{(n-1)}P_r = 504\)
\(\displaystyle (n-1)-r+1 = n-3 \) the last factor
\(\displaystyle n-n-r=-3\)
\(\displaystyle r=3 \)

so \(\displaystyle ^{(n-1)}P_r = ^{(n-1)}P_3 = 504\)
\(\displaystyle ^{(n-1)}P_3 = ^9P_3 = 504\)

so \(\displaystyle n-1 = 9\)
\(\displaystyle n = 10\)
\(\displaystyle 504n = \dfrac{n!}{(n - 4)!} = n(n - 1)(n - 2)(n - 3) \implies\)

\(\displaystyle (n - 1)(n^2 - 5n + 6) - 504 = 0 \implies n^3 - 6n^2 + 11n - 510 = 0.\)

There is a formula for solving cubics, but it is ugly.

The integer root theorem says that if that equation has an integer zero, it is an integer factor of 510. And we are only interested in positive zeroes.

\(\displaystyle 510 = 51 * 10 = 1 * 2 * 3 * 5 * 17.\)

Furthermore, we can easily see that n > 2 * 3 = 6 because

\(\displaystyle 6^3 - 6 * 6^2 + 11 * 6 - 510 = 6^3 - 6^3 + 66 - 510 << 0.\) And

\(\displaystyle (17^3 - 6 * 17^2 + 17 * 11 - 510 = 4913 - 1734 + 187) - 500 >> 0.\)

Thus, the only candidates left are 10 and 15.

\(\displaystyle 10^3 - 6 * 10^2 + 11 * 10 - 510 = 1000 - 600 + 110 - 510 = 400 + 110 - 510 = 0.\)

The integer root theorem can let you solve cubics of the form

\(\displaystyle x^3 + bx^2 + cx + d = 0\) if there is an integer root.

EDIT: Note that the integer root theorem does NOT guarantee that an integer root exists.

 
Last edited:
\(\displaystyle 504n = \dfrac{n!}{(n - 4)!} = n(n - 1)(n - 2)(n - 3) \implies\)

\(\displaystyle (n - 1)(n^2 - 5n + 6) - 504 = 0 \implies n^3 - 6n^2 + 11n - 510 = 0.\)

There is a formula for solving cubics, but it is ugly.

The integer root theorem says that if that equation has an integer zero, it is an integer factor of 510. And we are only interested in positive zeroes.

\(\displaystyle 510 = 51 * 10 = 1 * 2 * 3 * 5 * 17.\)

Furthermore, we can easily see that n > 2 * 3 = 6 because

\(\displaystyle 6^3 - 6 * 6^2 + 11 * 6 - 510 = 6^3 - 6^3 + 66 - 510 << 0.\) And

\(\displaystyle 17^3 - 6 * 17^2 + 17 * 11 - 510 = 4913 - 1734 + 187 - 500 >> 0.\)

Thus, the only candidates left are 10 and 15.

\(\displaystyle 10^3 - 6 * 10^2 + 11 * 10 - 510 = 1000 - 600 + 110 - 510 = 400 + 110 - 510 = 0.\)

The integer root theorem can let you solve cubics of the form

\(\displaystyle x^3 + bx^2 + cx + d = 0\) if there is an integer root.

EDIT: Note that the integer root theorem does NOT guarantee that an integer root exists.


There is a formula for solving cubics, but it is ugly.

:) this line made me remember a vision I saw yesterday about mathematics, that there're artistic ways to deal with math and unnecessary pressures ways to study it. I think it's like arithmetic and algebra, we can't use arithmetic ways to deal with advanced math but we use flexible algebra ways.
 
A small mistake in a number can change the whole problem
This is why I check my steps, as I go. I make more than enough silly mistakes, so checking what I've copied down (and each step after that) saves me a lot of time. :cool:
 
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