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In a party every person shakes hands with every other person. If there are 105 hands shakes, find the number of person in the party.
Answer & Solution
Correct Answer:
Option
A
Let n be the number of persons in the party
Number of hands shake = 105
Total number of hands shake is given by nC2 Now,
According to the question, $$\eqalign{ & ^n{{\text{C}}_2} = 105 \cr & {\text{or, }}\frac{{n!}}{{2! \times \left( {n - 2} \right)!}} = 105 \cr & {\text{or, }}\frac{{n \times \left( {n - 1} \right)}}{2} = 105 \cr & {\text{or, }}{n^2} - n = 210 \cr & {\text{or, }}{n^2} - n - 210 = 0 \cr & {\text{or, }}n = 15,\, - 14 \cr} $$ But, we cannot take negative value of n So, n = 15
i.e. number of persons in the party = 15
Number of hands shake = 105
Total number of hands shake is given by nC2 Now,
According to the question, $$\eqalign{ & ^n{{\text{C}}_2} = 105 \cr & {\text{or, }}\frac{{n!}}{{2! \times \left( {n - 2} \right)!}} = 105 \cr & {\text{or, }}\frac{{n \times \left( {n - 1} \right)}}{2} = 105 \cr & {\text{or, }}{n^2} - n = 210 \cr & {\text{or, }}{n^2} - n - 210 = 0 \cr & {\text{or, }}n = 15,\, - 14 \cr} $$ But, we cannot take negative value of n So, n = 15
i.e. number of persons in the party = 15
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LoginnCr = n!/[r!*(n-r)!]
So,
nC2 = n!/[2!*(n-2)!] as r = 2.
plz explain.