Victor has 100 trading cards, each with a distinct power level between 101 and 200 inclusive. Whenever he heads out, he always randomly selects 21 cards to bring with him just in case he meets a fellow collector. Prove that no matter which 21 cards he brings, Victor will always be able to select 4 of those cards that exhibit the following property: Let the average power level of all 4 cards bep. The cards can be split into two pairs, each of which also has an average power level of p.

Victor will always be able to select 4 of those cards with the following property

Explanation:

Number of trading cards = 100

victor selects 21 cards

let the 4 cards be labelled : A,B,C and D

The average power level of : A,B,C,D = ( A + B + C + D )/ 4 = P

let the two pairs be : ( A + B ) and ( C + D )

note: average power of each pair = P  and this shows that

( A + B ) = ( C + D ) for Victor to select 4 cards out of the 21 cards that exhibit the same property

we have to check out the possible choices of two cards out of 21 cards yield distinct sums.

= C(21,2)=(21x20)/2 = 210.

from the question the number of distinct sums that can be created using 101  through 200 is < 210 .

hence it is impossible to get 210 distinct sums therefore Victor will always be able to select 4 of those cards

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j