Permutations????

Question

4) Show that 6P4 = 6(5P3)
5) Show that nP5 - nP4 = (n-5)nP4
6) Solve for n: nP5 = 14(nP4)


How do I do these?

Thanks in advance

Answer 1

In general, nPk = n! / (n-k)!

So you just have to plug in your values and show how to get to the other value...

PROBLEM 4:
6P4 = 6! / (6-4)!

Pull out a 6 from the numerator:
6 * 5!
--------
(6-4)!

(6-4)! is the same as (5-3)! --> they are both 2! so replace it:

6 * 5!
-------
(5-3)!

Now look at 5! / (5-3)! --> it's the same as 5P3

6 * 5P3
= 6(5P3)
Done!


PROBLEM 5:
nP5 - nP4

.. n! ...... n!
------- - -------
(n-5)! ..(n-4)!

Notice that (n-5)! = (n-5)(n-6)(n-7) ... (2)(1)
And (n-4)! = (n-4)(n-5)(n-6)(n-7) ... (2)(1)

So if you multiply the left by (n-4) on top and bottom we will have:
(n-4) n! ......... n!
-------------- - -------
(n-4)(n-5)! ..(n-4)!

The left numerator can be changed to (n-4)!
(n-4) n! .... n!
---------- - -------
(n-4)! ..... (n-4)!

Now subtract because we have common denominators:
(n-4) n! - n!
---------------
.... (n-4)!

Pull out the common n! in the numerator:
n! ((n-4) - 1)
-----------------
.... (n-4)!

Simplify the item in parentheses:
n! (n-5)
-----------
.. (n-4)!

Now notice you have n! / (n-4)! which is the same as nP4
(n-5) nP4
Done!


PROBLEM 6:
nP5 = 14(nP4)

.. n! .......14 n!
-------- = -------
(n-5)! ... (n-4)!

Multiply the left hand side by (n-4) again so you get the same denominator:
(n-4) n! .. 14 n!
---------- = -------
(n-4)! ...... (n-4)!

So you can drop the denominators (multipy both sides by (n-4)!
(n-4) n! = 14 n!

Again drop the n! (divide both sides by n!)
n-4 = 14

Add 4 to both sides:
n = 18

So 18P5 = 14(18P4)
and n = 18