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Q. In a class of 30 students, each student shakes hands with every other student exactly once. How many handshakes occur in total?
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[tex]\huge\mathcal{\fcolorbox{orange}{azure}{\pink{➳ꪖꪀᦓ᭙ꫀ᥅ }}}[/tex]
To find the number of handshakes, we can use the formula [tex]\frac{n(n-1)}{2}[/tex],
where [tex]n[/tex] is the number of people in the group.
In this case, [tex]n = 30[/tex], so we can substitute that value into the formula to get [tex]\frac{30(30-1)}{2} = 435[/tex].
This formula works because each person shakes hands with [tex]n-1[/tex] other people, but we count each handshake twice (once for each person involved), so we need to divide the total number of handshakes by 2 to get the unique number of handshakes.
There are [tex]\frac{n(n-1)}{2}[/tex] unique handshakes in a group of [tex]n[/tex] people. In this case, [tex]n = 30[/tex], so there are [tex]\frac{30(30-1)}{2} = 435[/tex] handshakes in total.
[tex]\huge\mathcal{\fcolorbox{orange}{azure}{\pink{➳Tнапкчѳц }}}[/tex]
Answer:
To find the total number of handshakes, we can use the formula for the sum of the first n-1 positive integers, which is given by:
Total handshakes = (n-1) + (n-2) + (n-3) + ... + 3 + 2 + 1
In this case, n represents the number of students in the class, which is 30. Therefore, we can substitute n = 30 into the formula:
Total handshakes = (30-1) + (30-2) + (30-3) + ... + 3 + 2 + 1
Simplifying this expression, we get:
Total handshakes = 29 + 28 + 27 + ... + 3 + 2 + 1
To calculate the sum, we can use the formula for the sum of an arithmetic series, which is given by:
Sum = (n/2)(first term + last term)
In this case, the first term is 1 and the last term is 29, and n is the number of terms, which is 29.
[tex] \\ Sum = \bigg(\frac{29}{2} \bigg)(1 + 29) \\ \bigg(\frac{29}{2} \bigg) \times (30) = 450[/tex]
Therefore, the total number of handshakes that occur in the class of 30 students is 450.
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