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junior college 1 | H1 Maths
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Hi, my senior needs help with this question. Thank you :)
The answer is 5,311,735
Date Posted: 3 years ago
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The 26-term series is to be written with either -5 or +8. The sum of the terms must be 0.
Notice that for the terms to cancel out and become 0, we need eight "-5"s to negate every five "+8"s. That's a total of thirteen (13) integers.
(Because the common multiple of 5 and 8 is 40. So 8 x 5 + 5 x 8 = -40 + 40 = 0)
For a 26-term series, 26 ÷ 13 = 2
So we'll need 2 sets of the above. i.e sixteen "-5"s and ten "+8"s.
One such example of a series would be :
-5 + 8 + 8 + 8 - 5 - 5 + 8 + 8 - 5 - 5 - 5 + 8 - 5 - 5 + 8 - 5 - 5 - 5 + 8 - 5 + 8 - 5 - 5 - 5 + 8 - 5 = 0
Now these integers can be rearranged and they would still give a sum of 0.
①What we can do is to choose ten positions out of 26 for the "+8"s. The "-5"s will then be put into the remaining sixteen positions.
Number of ways to choose ten positions
= 26C10 = 26! ÷ 10! ÷ (26 - 10)!
= 26! ÷ 10! ÷ 16!
= 5 311 735
Since there are 5311 735 ways to choose this , there are 5 311 735 ways to write the series.
②Alternatively, choose sixteen positions out of 26 for the "-5"s. The "+8"s will then be put into the remaining ten positions.
Number of ways to choose sixteen positions
= 26C16 = 26! ÷ 16! ÷ (26 - 16)!
= 26! ÷ 16! ÷ 10!
= 5 311 735
③ Alternatively,
Number of unique ways to arrange 26 terms = 26!
But this only works if each of the 26 terms are unique and different.
Since the sixteen "-5"s are identical to each other and are indistinguishable (likewise for the ten "+8"s), we need to divide by the number of permutations for each of these groups to get rid of repeated cases.
Number of permutations for the "-5"s if they were unique = 16!
Number of permutations for the "+8"s if they were unique = 10!
So, number of unique ways to permutate the 26 terms = 26! ÷ 10! ÷ 16! = 5 311 735
Notice that for the terms to cancel out and become 0, we need eight "-5"s to negate every five "+8"s. That's a total of thirteen (13) integers.
(Because the common multiple of 5 and 8 is 40. So 8 x 5 + 5 x 8 = -40 + 40 = 0)
For a 26-term series, 26 ÷ 13 = 2
So we'll need 2 sets of the above. i.e sixteen "-5"s and ten "+8"s.
One such example of a series would be :
-5 + 8 + 8 + 8 - 5 - 5 + 8 + 8 - 5 - 5 - 5 + 8 - 5 - 5 + 8 - 5 - 5 - 5 + 8 - 5 + 8 - 5 - 5 - 5 + 8 - 5 = 0
Now these integers can be rearranged and they would still give a sum of 0.
①What we can do is to choose ten positions out of 26 for the "+8"s. The "-5"s will then be put into the remaining sixteen positions.
Number of ways to choose ten positions
= 26C10 = 26! ÷ 10! ÷ (26 - 10)!
= 26! ÷ 10! ÷ 16!
= 5 311 735
Since there are 5311 735 ways to choose this , there are 5 311 735 ways to write the series.
②Alternatively, choose sixteen positions out of 26 for the "-5"s. The "+8"s will then be put into the remaining ten positions.
Number of ways to choose sixteen positions
= 26C16 = 26! ÷ 16! ÷ (26 - 16)!
= 26! ÷ 16! ÷ 10!
= 5 311 735
③ Alternatively,
Number of unique ways to arrange 26 terms = 26!
But this only works if each of the 26 terms are unique and different.
Since the sixteen "-5"s are identical to each other and are indistinguishable (likewise for the ten "+8"s), we need to divide by the number of permutations for each of these groups to get rid of repeated cases.
Number of permutations for the "-5"s if they were unique = 16!
Number of permutations for the "+8"s if they were unique = 10!
So, number of unique ways to permutate the 26 terms = 26! ÷ 10! ÷ 16! = 5 311 735
Thabk you so much :)
Welcome. There is a small typo in the earlier part of the working.
It should be -8 x 5 + 5 x 8 = -40 + 40 = 0
There was a missing negative sign on the 8.
It should be -8 x 5 + 5 x 8 = -40 + 40 = 0
There was a missing negative sign on the 8.
Thank you so much :)
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