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

the first, second, and fifth options

emily ruth's annual premium is $1058.50

step-by-step explanation:

emily ruth is a female of 27 years age purchasing 20 year endowment insurance with face value $25000.

from the given table,

for females, the age of a male 3 years younger should be used.

so, we have to check for 24 year male for 20 year endowment.

it is 42.34 for every $1000 face value.

for a face value of $25000,

it is 25 * 42.34 = $1058.50

∴ emily ruth's annual premium is $1058.50