6-letter words from (a,b,c,d,e,f) using "a" twice "b" twice and no other letter more than once??

Question

How many six-letter words can be made from (a,b,c,d,e,f) using "a" twice "b" twice and no other letter more than once?

NOTE: THEY DONT HAVE TO BE REAL WORDS.

How about using "a" 3 times and "b" once and no other letter more than once?

How abt using "a" 3 times and "b" twice and no other letter more than once?

How abt using exactly two "a" and one "b" and no other letter more than once?

How abt in (a, b, c, d, e, f, g, h, 1, 2, 3, 4), how many subsets contain all four numerals? How many contain exactly two numerals? How many contain exactly 3 letters and two numerals?

I need a general idea how to do the problems. Thank you!

Answer 1

First, a clarification: If 'a', 'b' MUST be used twice, the outcome is different, than if 'a' and 'b' MAY be used twice. I assume 'a' and 'b' must be used twice (cannot be more or less than twice).

I'm glad you asked you want to know the 'general idea of solving these types of problems', rather than numeric answers for many variants you've posed in the question. Yes, the essence of the matter is the underlying principles.

Think of the 6-letter word to be a 6 placeholders of one letter each. Use the permutation method on how many ways can the first place be filled ? the second place ? and so on.

- First 'a' needs to be in the word: 6 possible places.
- Second 'a' needs to be in the word: 5 possible places.
- Similarly for the 2 'b's - 4, 3 ways of using them.
- Now there are two remaining placeholders; but {c,d,e,f} are
available to fill them: 4 and 3 possible ways.

Hence the total possible ways (permutations) of creating such 6-letter words are: 6x5x4x3x4x3 = 4320 words.

Now crack the rest of the problem yourself. Good luck.

Answer 2

First, if you have a number of objects that you want to list, you can find the number of ways this can be done if repetition constraints are added by:
(number of objects)! / [(repetition1)! * (repetition2)! ...]

In the first case, the required number is 6!/[2!*2!] = 180 ways.

! is read factorial and k! means 1*2*...k, i.e., the product of numbers from 1 up to k.

You can likewise do the rest of the problems.

Second, the final problem.
If you have n objects that you want to choose k from, use the combination formula:
C(n,k) = n!/[k! *(n-k)!]

Applying this leads to the fact that there are C(12-4,5-4) = C(8,1) = 8 subsets of size 5 containing four numerals,
C(12-4,6-4) = C(8,2) = 28 subsets of size 6 having four num.
etc... until C(12-4,12-4) = 1 contian four numerals.
So there are the sum of all those previous numbers of subsets that contain four numerals.

Those that contain exactly 3 letters and two numbers are in number equal to C(8,3)*C(4,2) = 56*6 = 336.

Those subsets that contain exactly two numberals can be treated as those that contain four.

Answer 3

Assume second a be g, second b be h, then arrangement N of 8 by 6 = 8!/2!, now exclude
the same situations /ga and ag/ and /bh and hb/, then N=(8!/2!)/2/2 =7!

Answer 4

fabbed
babead
babade
cabbed
bacade
beddac
acedab
dabbed
dabace
debacaa