Here I am on the last section of my "Permutations and Combinations" topic with this only one example dealing with an unique kind of problem as quoted below:
Now my problem quotes as follows:Example: Three letters are selected at random from the letters of the word BIOLOGY.
Find the total number of selections.
The answer is not as you might expect.
Because there are two letters O, you need to find the number of selections with
1) no letters O
2) one letter O
3) two letters O
and then add these together.
1) Number of selections with no letter O (ex. B, L and Y)
= number of ways to choose three letters from B, I, L, G, Y
= = 10
2) Number of selections with one letter O (ex. O, B and L)
= number of ways to choose two letters from B, I, L, G, Y
= = 10
3) Number of selections with two letters O (ex. O, O and B)
= number of ways to choose one letter from B, I, L, G, Y
= 5
Therefore, total number of selections = 10 + 10 + 5 = 25
In a mixed pack of coloured light bulbs there are three red bulbs, one yellow bulb, one blue bulb and one green bulb.
Four bulbs are selected at random from the pack.
How many different selections are possible?
Ans: 7
That means if I represent all the 6 bulbs with their corresponding color letters as in the set below,
How many combinations of 4 letters from the 6 letters, where 3 letters R is repeated, can I get?
{ R R R Y B G }
But I still have trouble applying the example solution there . I need help please ! Thanks.
Thanks Plato.
But I can confirmed the answer is indeed 7,
1. R Y B G
2. R R Y B
3. R R Y G
4. R R B G
5. R R R Y
6. R R R B
7. R R R G
As you can see, there can't be better combinations than these.
I just expect someone to give a general and clear worked out solution or formula (or like the one given in the first example provided you get the right answer from it). And also because in my textbooks there a lack of references on Combination from objects that are not distinct (that include alike/repetitive objects).
But just now Considering the 7 combinations just listed above, I think I can see clearly the clue about how to tackle it step by step,
For 4 Selections:
1)
without any repetitions of R there are R, Y, B, G
= = 1
2-4)
with 1 repetition of R there are now 2 R (fixed) with remaining 3 distinct objects (Y B G) to be selected in 2 spaces, ex. (R,R, ... , ...)
= = 3
5-7)
with 2 repetitions of R there are now 3 R (fixed) with remaining 3 distinct objects (Y B G) to be selected in 1 space, ex. (R,R,R, ...)
= = 3
Finally adding all combinations together, 1 + 3 + 3 = 7.
All Credits goes to Sir Plato for guidance.
Again thanks.
Well that is categorically not what you asked.
You asked "How many combinations of 4 letters from the 6 letters, where 3 letters R is repeated, can I get?"
Look at your #'s 1, 2, 3 & 4, only contain two R's.
Moreover, you used the word combinations which does not imply order.
Combinations are about content so as a better word is selections.
From the possible choices $R,~Y,~G,~B$ how many selections of four are possible that have three R's.
I answered that: $3$.
If you mean, How many ways are there to rearrange the string $RRRG$ then the answer is four.
If you mean "How many 4-strings made with three R's one of G, Y, or B?" then the answer is twelve.
It does not appear that you are clear on what is being asked?
I am very sorry
From:
"How many combinations of 4 letters from the 6 letters, where 3 letters R is repeated, can I get?"
I meant the 3 Letters R repeated in the 6 letters of the set { R R R Y B G } from which to make the selections instead of the selections itself.
Hope you can excuse my confusing English also.
Little mistake make huge problem, I apologies