How many different integers can be expressed as the sum of three distinct numbers from the set 13 10 23

8C3=8!/5!*3!=56
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please some0ne provide the correct solution
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ans is 54
8C3=56
56-2=54
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plz explain this
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total number of selection of 3 number out of 8 number is=8c3=56
10+28+33=10+13+48
10+13+43=10+23+33
the answer is=56-2=54
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total possible integers=8C3. but there are 6 integers repeating twice. so we have to exclude 6. so total different integers are 56-6=50.
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Answer is 32.
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please give a proper valid reason for the answer 54?
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plzz explain it properly.. why 56-2?
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its not termed as all the no i belong to the set so the no is (0to9)
so there is 2 option the sum is either 2 digit no. or 3 digit no.
for the 2 digit no 9c1*9c1=81
for the 3 digit no 9c1*9c2=324
so there are 324+81=405 distinct no possible
10 years agoHelpfull: Yes(0) No(2)
answer will be 54 because by 8C3 be choose set of intergers but out of this (13,10,23) & (23,10,33) are similar so for different set we get 56-2 that is 54
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15 is the sol since the comm diff is 7 their sum also has a comm diff 7the min sium is 30 and max sum is 135 the remaining sums are multiple of 7 in between 30 and 135 . there fore the ans is (135-30)/7 = 105/7 = 15
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How many different integers can be expressed as the sum of three distinct numbers from the set{13,10,23,28,33,36,43,48}?