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Question Number 203788 by BaliramKumar last updated on 28/Jan/24
The sum of all possible products taken two at   a time  out of the numbers ±1, ±2, ±3, ±4 is  (a) 0           (b) −30           (c) 30           (d) 55
Thesumofallpossibleproductstakentwoatatimeoutofthenumbers±1,±2,±3,±4is(a)0(b)30(c)30(d)55
Answered by mr W last updated on 28/Jan/24
to take two from 8 there are   C_8 ^2 =((8×7)/2)=28 possibilities    group 1: (+1,+2, +3, +4)  group 2: (−1,−2, −3, −4)    when we take two numbers, there  are following cases:    case 1:   both numbers from group 1  C_4 ^2 =((4×3)/2)=6 possibilities  1×2+1×3+1×4+2×3+2×4+3×4=35    case 2:   both numbers from group 2  also 6 possibilities  sum is also 35.    case 3:   one number from group 1 and the  other number from group 2  4×4=16 possibilities  (−1−2−3−4)×(1+2+3+4)=−100    summary:  6+6+16=28 possibilities ✓  sum=35+35−100=−30 ✓  ⇒answer (b) is correct
totaketwofrom8thereareC82=8×72=28possibilitiesgroup1:(+1,+2,+3,+4)group2:(1,2,3,4)whenwetaketwonumbers,therearefollowingcases:case1:bothnumbersfromgroup1C42=4×32=6possibilities1×2+1×3+1×4+2×3+2×4+3×4=35case2:bothnumbersfromgroup2also6possibilitiessumisalso35.case3:onenumberfromgroup1andtheothernumberfromgroup24×4=16possibilities(1234)×(1+2+3+4)=100summary:6+6+16=28possibilitiessum=35+35100=30answer(b)iscorrect
Commented by mr W last updated on 28/Jan/24
i think we can generalize:  let′s say we take two numbers out  of ±1, ±2, ±3, ..., ±n  the sum of all possible products is  −(1^2 +2^2 +3^2 +...+n^2 )=−((n(n+1)(2n+1))/6)  e.g. n=4: −((4×5×9)/6)=−30 ✓
ithinkwecangeneralize:letssaywetaketwonumbersoutof±1,±2,±3,,±nthesumofallpossibleproductsis(12+22+32++n2)=n(n+1)(2n+1)6e.g.n=4:4×5×96=30
Commented by BaliramKumar last updated on 28/Jan/24
Thanks Sir
ThanksSir

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