3 thoughts on “11Ext: Topic 5: Permutations & Combinations QOB”
i dont know how to do question 17 exercise 10 F! it’s so hard i cant even think of any way approaching it!
Yes, this is one of the hardest questions in the Perms & Combs section, let me try to explain.
Questions where you are asked for things to be “not” together, are often easier to work out “are” together, and take away from total number with no restrictions, this is the approach I will take.
Think of the problem as this;
How many 8 letter words can be formed from the letters A,A,B,B,C,C,D,D, so that no word has a double letter, essentially the same problem, I’m just using letters instead of shirts!
1. Total words, no restrictions = 8!/(2!2!2!2!) = 2520
2. Need to take away the words with a double letter
Total words with a double letter (e.g. AA) we treat the double letter as if it is one letter as the A and A must stay connected, so question becomes how many ways can we arrange the letters (AA),B,B,C,C,D,D , 7 letters in total
Total words with a double letter = 4 X 7!/(2!2!2!) = 2520
times 4 as there are 4 letters that could be doubled.
3. Problem is when we worked out the numbers of words with one double letter, it included the words with two double letters, think of a venn diagram
| words with AA |
___________________________________
| AABCDBCD | AABBCDCD | BBACDACD |
| CDAABDCB | CDBBAACD | CADBBCDA |
| etc | etc | etc |
|___________|__________|___________|
| words with BB |
The set with words with AA intersects with the set with words with BB, so when we take away these two sets, the set with words with both AA and BB has been taken away twice, so we need to add these possibilities back in
Total words with two double letters (e.g AA and BB), question is how many words can be formed from the letters (AA),(BB),C,C,D,D , 6 letters in total
Total words with two double letters = 4C2 X 6!/(2!2!) = 1080
times 4C2 as there are 4 letters that could be doubled, we need to choose two.
4. When we added back in the set of letters with two doubles, we included the set with three letters twice, just like the last case but this time because we have take away the possiblilities
total words with three double letters (e.g. AA , BB and CC), so how many words can be formed from the letters (AA),(BB),(CC),D,D , 5 letters in total
Total words with three double letters = 4C3 X 5!/2! = 240
times 4C3 as there are 4 letters that could be doubled, we need to choose three
5. Now when we took away the set of words with three double letters, we now have double counted the words with four double letters, and they have to be added back in
Total words with four double letters (e.g AA , BB , CC and DD), how many words can be formed from the letters (AA) , (BB) , (CC) , (DD) , 4 letters in total
total words with four double letters = 4! = 24
6. Putting it all together
Total Words = 2520 – 2520 + 1080 – 240 + 24 = 864 !!!
I hope that makessome sort of sense.
Thanks a lot for the detailed explanation! i get it now! no one has ever explained it to me just like the way you did! Thanks again
i dont know how to do question 17 exercise 10 F! it’s so hard i cant even think of any way approaching it!
Yes, this is one of the hardest questions in the Perms & Combs section, let me try to explain.
Questions where you are asked for things to be “not” together, are often easier to work out “are” together, and take away from total number with no restrictions, this is the approach I will take.
Think of the problem as this;
How many 8 letter words can be formed from the letters A,A,B,B,C,C,D,D, so that no word has a double letter, essentially the same problem, I’m just using letters instead of shirts!
1. Total words, no restrictions = 8!/(2!2!2!2!) = 2520
2. Need to take away the words with a double letter
Total words with a double letter (e.g. AA) we treat the double letter as if it is one letter as the A and A must stay connected, so question becomes how many ways can we arrange the letters (AA),B,B,C,C,D,D , 7 letters in total
Total words with a double letter = 4 X 7!/(2!2!2!) = 2520
times 4 as there are 4 letters that could be doubled.
3. Problem is when we worked out the numbers of words with one double letter, it included the words with two double letters, think of a venn diagram
| words with AA |
___________________________________
| AABCDBCD | AABBCDCD | BBACDACD |
| CDAABDCB | CDBBAACD | CADBBCDA |
| etc | etc | etc |
|___________|__________|___________|
| words with BB |
The set with words with AA intersects with the set with words with BB, so when we take away these two sets, the set with words with both AA and BB has been taken away twice, so we need to add these possibilities back in
Total words with two double letters (e.g AA and BB), question is how many words can be formed from the letters (AA),(BB),C,C,D,D , 6 letters in total
Total words with two double letters = 4C2 X 6!/(2!2!) = 1080
times 4C2 as there are 4 letters that could be doubled, we need to choose two.
4. When we added back in the set of letters with two doubles, we included the set with three letters twice, just like the last case but this time because we have take away the possiblilities
total words with three double letters (e.g. AA , BB and CC), so how many words can be formed from the letters (AA),(BB),(CC),D,D , 5 letters in total
Total words with three double letters = 4C3 X 5!/2! = 240
times 4C3 as there are 4 letters that could be doubled, we need to choose three
5. Now when we took away the set of words with three double letters, we now have double counted the words with four double letters, and they have to be added back in
Total words with four double letters (e.g AA , BB , CC and DD), how many words can be formed from the letters (AA) , (BB) , (CC) , (DD) , 4 letters in total
total words with four double letters = 4! = 24
6. Putting it all together
Total Words = 2520 – 2520 + 1080 – 240 + 24 = 864 !!!
I hope that makessome sort of sense.
Thanks a lot for the detailed explanation! i get it now! no one has ever explained it to me just like the way you did! Thanks again