How many people will be needed, so that if every people shook their hands with each other, there will be 91 handshakes?
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See every people will do handshake with 1 of his hand and so total handshakes that took place we can simply use this formula n(n-1)/2 as the no of handshakes are in AP where n = no.of persons
so 91 x 2 = n² - n
⇒ n² - n - 182 = 0
⇒n² - 14n + 13n - 182 = 0
⇒ n(n - 14) + 13(n - 14) = 0
⇒(n - 14)(n + 13) = 0
so n = -13 or 14 so no. of persons can't be negative so no of persons is 14 ANSWER
so 91 x 2 = n² - n
⇒ n² - n - 182 = 0
⇒n² - 14n + 13n - 182 = 0
⇒ n(n - 14) + 13(n - 14) = 0
⇒(n - 14)(n + 13) = 0
so n = -13 or 14 so no. of persons can't be negative so no of persons is 14 ANSWER
Anonymous:
hope it helps
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Nice photo you have there.
We have n persons. Handshake involves selection of two persons. Order of selection of these two is not important. That is, it is the combination of the two persons, makes one handshake.
Hence there are nC₂ hand shakes in a group of n persons.
nC₂ = n! / 2! (n-2)! = n(n-1)/2 = 91
n (n - 1) = 2 * 7 * 13
n (n - 1) = 14 * 13
hence, n = 14
We have n persons. Handshake involves selection of two persons. Order of selection of these two is not important. That is, it is the combination of the two persons, makes one handshake.
Hence there are nC₂ hand shakes in a group of n persons.
nC₂ = n! / 2! (n-2)! = n(n-1)/2 = 91
n (n - 1) = 2 * 7 * 13
n (n - 1) = 14 * 13
hence, n = 14
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