# Math Help - Yet another (two) permutations!

1. ## Yet another (two) permutations!

Hello,

I have recieved great response for my previous questions regarding permutations and I am here once again for some more help.

1) Fran is working on a word puzzle and is looking for four-digit "scrambles" from the clue word calculate.

a) How many of the possible four-letter scrambles contain four different letters?

b) How many contain two a's and one other pair of identical letters?

c) How many scrambles consist of any two pairs of identical letters?

d) What possiblities have you not yet taken into account? Find the number of scrambles for each of these cases.

e) What is the total number of four-letter scrambels taking all cases into account?

2) How many numbers can be formed using all of the digits 1, 2, 3, 4, 5, 6, and 7 if the odd digits must be in ascending order and the even digits in descending order?

2. Originally Posted by NineZeroFive
1) Fran is working on a word puzzle and is looking for four-digit "scrambles" from the clue word calculate.
a) How many of the possible four-letter scrambles contain four different letters?
b) How many contain two a's and one other pair of identical letters?
c) How many scrambles consist of any two pairs of identical letters?
d) What possiblities have you not yet taken into account? Find the number of scrambles for each of these cases.
e) What is the total number of four-letter scrambels taking all cases into account?

2) How many numbers can be formed using all of the digits 1, 2, 3, 4, 5, 6, and 7 if the odd digits must be in ascending order and the even digits in descending order?
I am not sure that I understand the idea of “a scramble”. Does the (a) question mean “How many of the possible four-letter “words” contain four different letters from the set of letters in CALCULATE?
If that is correct, the answer is: $_6 P_4 = \frac{{6!}}{{\left( {6 - 4} \right)!}} = 6 \cdot 5 \cdot 4 \cdot 3$
Because, there are just there are only six different letters.

Frankly I find the next three parts very poorly put.
I can help with part e. Consider three cases: no repeated letters (part a), exactly one pair of repeated, and two pairs of repeats.
Combinations: $\left( {_n C_k } \right) = \frac{{n!}}{{\left( {n - k} \right)!\left( {k!} \right)}}$

Exactly one pair of repeated: $\left( {_3 C_1 } \right)\left( {_5 C_2 } \right)\frac{{4!}}{{2!}}$. Select the repeated one from three. Then select two from the other five to be different. Now arrange these.
Two pairs of repeated letters: $\left( {_3 C_2 } \right)\frac{{4!}}{{\left( {2!} \right)^2 }}$.

So the answer is $\left( {_7 C_4 } \right)$.