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Saumyojit

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In how many ways can a selection be made of 5 letters out of 5As ,4 Bs, 3Cs, 2 Ds and 1E?

I could not find any approach
 
Saumyojit said:
In how many ways can a selection be made of 5 letters out of 5As ,4 Bs, 3Cs, 2 Ds and 1E?

I could not find any approach
Click to expand...
Start listing the choices:

AAAAA, AAAAB, AAAAC.....

What does that tell you?
 
Subhotosh Khan said:
Start listing the choices:

AAAAA, AAAAB, AAAAC.....

What does that tell you?
Click to expand...
All 5 letters from same alphabet - 1 way (A)
4 being same letter+1 different: 2∗4C1=8 ways
3 same ,1 different, 1 another different: 3∗4C2=18 ways
2 same, and rest all different : 4∗4C3=16 ways
All five different: 1 way
3 of one type and 2 of 1 type : 3∗3C1=9 ways


1 of one type, 2 of 2nd type, 2 of 3rd type : From A,B,C,D -> I can select anyone out of four : Then out of three any one -> 4c2 . Suppose A ,b is selected then it is like this : AA BB

Now i am left with C,D,E --> 3c1 . SO , 4c2 * 3c1 .


Edit: I think this is the problemn : 4c2 should be used instead of 4c1 *3c1 as the latter expression is over counting . I just did that with that one eg. ( intuition)
I understood it .

Explanation of 4c1 * 3c1 : A or B or C or D * (B or C or D) OR (A or C or D) OR (A or B or D) or (A or B or C) . SO it leads to double .


DD BB E

BB DD E .
 
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Saumyojit said:
In how many ways can a selection be made of 5 letters out of 5As ,4 Bs, 3Cs, 2 Ds and 1E?
I could not find any approach
Click to expand...
Do not over-think this question. All five A's are identical. That is true of the other denominations.
So you can have 5A's. 4A's1B, 4A's 1C, 4A's 1D, & 4A's 1E. OR 3A's 2B', 3A's 2C', & 3A's 2D's.
Can you complete the list in which the last entry is 1A, 1B, 1C, 1D & 1E. ??
 
I don’t get why my reply to this get removed and posted as a new thread… First of all, if no letters can repeat, you will have p5,5 = 120 ways to form a word. And Second, if they do repeat, in how many ways they repeat and for each way the letters repeat how many ways is there to form a 5 letter word. If we view the p5,5 format as 1+1+1+1+1, then there is obviously many 3+2, 2+2+1 formats, for corresponding example: BBBCC, CCDDE…
 
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Here’s a list of analyzed formats: 5+0, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1 for example:

5+0: AAAAA

4+1: AAAAC, BBBBE, BBCBB

3+1+1: CCCAE

….
 
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Each format is a mini problem. Take the 3+2 format as an example: ABC can repeat 3 times and ABCD can repeat 2 times yet you can’t have AAA+AA. So for 3A format there are BCD. C(3,1) to use for the other letter to repeat twice. Same as for 3B and 3C and in total you will have 9 sub formats. For each sub format, there are many ways to form a word, taking 3A2B as example: ABBAA, AABBA, AAABB, BBAAA, ABABA, AABAB, BABAA. So just this format you are looking at 63 events.
 
5+0, 4+1, 3+2 and 1+1+1+1+1 are the easiest format to calculate the arrangements: 5+0: 1 event, 4+1: 40 events, 3+2: 63 events, 1+1+1+1+1: 120 events. Each is unique word. And if you don’t find this complicated enough calculate the other formats for instance 3+1+1: ABC to choose from to form 3A 3B 3C format, then each has 4 other letters to place in front, in the middle or behind the 3 letters, which gives you 4 spaces to place, and then the last letter you will choose from the remaining 3 and place in the 5 spaces in front, in the middle or behind. Total will be 3*16*15 which is 720 events.
 
2+2+1, you will have 4 to choose from to form the first letter repeating twice, and 3 to choose from to form the second letter to repeat twice: A2B2, A2C2, A2D2, B2A2, B2C2, B2D2, C2A2, C2B2, C2D2, then D is not needed to start again for it is redundant in the set. So we examine this format, each sub format can form a four letter word such as ABBA, AABB, BBAA, ABAB, BABA. And for all 5 variations there are 5 spaces to place a letter which can be chosen from 3 remaining letters. So for this format you have 9*15 = 135 unique events.
 
I should as well do 2+1+1+1 for you: 2A, 2B, 2C, 2D to start, each can have a insert front, middle and behind as 3 spaces, choosing from the remaining 4 letters, then 4 spaces choosing from the remaining 3 letters, then 5 spaces choosing from the remaining 2 letters. 4*12*12*10=5760 events.

1+40+63+720+135+5760+120=6839 ways to form a 5 letter word.
 
I made a mistake at 2+2+1 format. 5 variations there are 5 spaces to place a letter which can be chosen from 3 remaining letters should be: 5*5*3, multiply that to 9 different 2+2 subformats there are 675 ways to form.


1+40+63+720+675+5760+120=7379 ways to form a 5 letter word.
 
sky2rain said:
I don’t get why my reply to this get removed and posted as a new thread…
Click to expand...
You posted it in the wrong thread. (This problem was in an image there, but the thread was about the other problem.)

I replied, telling you that, but my reply was removed and your post moved without explanation.
 
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