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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
$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{products}\:\mathrm{taken}\:\mathrm{two}\:\mathrm{at}\: \\ $$$$\mathrm{a}\:\mathrm{time}\:\:\mathrm{out}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}\:\pm\mathrm{1},\:\pm\mathrm{2},\:\pm\mathrm{3},\:\pm\mathrm{4}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:−\mathrm{30}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{30}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{55} \\ $$
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
$${to}\:{take}\:{two}\:{from}\:\mathrm{8}\:{there}\:{are}\: \\ $$$${C}_{\mathrm{8}} ^{\mathrm{2}} =\frac{\mathrm{8}×\mathrm{7}}{\mathrm{2}}=\mathrm{28}\:{possibilities} \\ $$$$ \\ $$$${group}\:\mathrm{1}:\:\left(+\mathrm{1},+\mathrm{2},\:+\mathrm{3},\:+\mathrm{4}\right) \\ $$$${group}\:\mathrm{2}:\:\left(−\mathrm{1},−\mathrm{2},\:−\mathrm{3},\:−\mathrm{4}\right) \\ $$$$ \\ $$$${when}\:{we}\:{take}\:{two}\:{numbers},\:{there} \\ $$$${are}\:{following}\:{cases}: \\ $$$$ \\ $$$${case}\:\mathrm{1}:\: \\ $$$${both}\:{numbers}\:{from}\:{group}\:\mathrm{1} \\ $$$${C}_{\mathrm{4}} ^{\mathrm{2}} =\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}}=\mathrm{6}\:{possibilities} \\ $$$$\mathrm{1}×\mathrm{2}+\mathrm{1}×\mathrm{3}+\mathrm{1}×\mathrm{4}+\mathrm{2}×\mathrm{3}+\mathrm{2}×\mathrm{4}+\mathrm{3}×\mathrm{4}=\mathrm{35} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\: \\ $$$${both}\:{numbers}\:{from}\:{group}\:\mathrm{2} \\ $$$${also}\:\mathrm{6}\:{possibilities} \\ $$$${sum}\:{is}\:{also}\:\mathrm{35}. \\ $$$$ \\ $$$${case}\:\mathrm{3}:\: \\ $$$${one}\:{number}\:{from}\:{group}\:\mathrm{1}\:{and}\:{the} \\ $$$${other}\:{number}\:{from}\:{group}\:\mathrm{2} \\ $$$$\mathrm{4}×\mathrm{4}=\mathrm{16}\:{possibilities} \\ $$$$\left(−\mathrm{1}−\mathrm{2}−\mathrm{3}−\mathrm{4}\right)×\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)=−\mathrm{100} \\ $$$$ \\ $$$${summary}: \\ $$$$\mathrm{6}+\mathrm{6}+\mathrm{16}=\mathrm{28}\:{possibilities}\:\checkmark \\ $$$${sum}=\mathrm{35}+\mathrm{35}−\mathrm{100}=−\mathrm{30}\:\checkmark \\ $$$$\Rightarrow{answer}\:\left({b}\right)\:{is}\:{correct} \\ $$
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 ✓
$${i}\:{think}\:{we}\:{can}\:{generalize}: \\ $$$${let}'{s}\:{say}\:{we}\:{take}\:{two}\:{numbers}\:{out} \\ $$$${of}\:\pm\mathrm{1},\:\pm\mathrm{2},\:\pm\mathrm{3},\:…,\:\pm{n} \\ $$$${the}\:{sum}\:{of}\:{all}\:{possible}\:{products}\:{is} \\ $$$$−\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} \right)=−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${e}.{g}.\:{n}=\mathrm{4}:\:−\frac{\mathrm{4}×\mathrm{5}×\mathrm{9}}{\mathrm{6}}=−\mathrm{30}\:\checkmark \\ $$
Commented by BaliramKumar last updated on 28/Jan/24
Thanks Sir
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$

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