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Sagot :
The number of different strings that can be made from the letters in ORONO using some or all of the letters is 63
What is the number of permutations in which n things can be arranged such that some groups are identical?
Suppose there are n items.
Suppose we have
[tex]i_1, i_2, ..., i_k[/tex]
sized groups of identical items.
Then the permutations of their arrangements is given as
[tex]\dfrac{n!}{i_1! \times i_2! \times ... \times i_k!}[/tex]
We're given the word ORONO.
- Case 1: One letter string:
3 ways: O, R, or N
- Case 2: Two letter string:
Subcase: Consists O:
5 ways: OR, RO, ON, NO, OO
Subcase: Doesn't consist O:
2 way: RN, NR
Total permutation under this case = 5+2 = 7
- Case 3: Three letter string:
Subcase: Consists 3 'O':
1 way: OOO
Subcase: Consists 2 'O':
6 ways: OOR, ROO, ORO, OON, NOO, ONO
Subcase: Consists 1 'O':
6 ways: ORN, ONR, RNO, RON, NOR, NRO (or that there are 3 distinct letters to be arranged, which can be done in 3! = 6 ways)
Subcase: Doesn't consist O:
0 ways as only R and N cannot form three letter string.
Total permutation under this case = 1+6+6= 13
- Case 4: 4 letter string:
Subcase: Consists 3 'O':
So, we've got {O,O,O,R,N}
3 'O's are mandatory, so fourth letter is either R or N:
O, O, O, R can arrange themselves in : [tex]4!/3! = 4 \: \rm ways[/tex] (as total 4 words but 3 are identical)
Similarly, O, O, O, N can arrange themselves in 4 ways,
So when 4 letter string from ORONO consists of 3 'O's, then total 4+4=8 distinct strings can be made.
Subcase: Consists 2 'O':
So, we've got {O,O,R,N}
The total permuations of these 4 letters, of which 2 are identital is:
[tex]\dfrac{4!}{2!} = 12[/tex]
Subcase: Consists 1 'O':
0 ways: Single 'O' and one-one R and N cannot form four letter string.
Subcase: Doesn't consist O:
0 ways as only R and N cannot form four letter string.
Total permutation under this case = 8+12 = 20
- Case 5: Five lettered string:
All 5 letters of ORONO would be used.
There are 3 identical objects, so total number of their permutation is:
5!/3! = 20
Thus, from all these cases, we conclude that:
Total permutations of string letters using some or all letters of ORONO is:
3+7+13+20+20 = 63
Thus, the number of different strings that can be made from the letters in ORONO using some or all of the letters is 63
Learn more about permutations here:
https://brainly.com/question/13443004
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