Proposed Problem

Discussion in 'Mathematics' started by AVDBA, Sep 21, 2019.

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1. AVDBANew commenter

A global leader of Educational Services has decided to form, worldwide, 15 commissions
to improve these services. N specialists were chosen to compose the commissions.
Each specialist must belong to two commissions and each pair of commissions must
have, exactly, one member in common.

Determine: a) N, the number of specialists; b) the number of specialists within
each commission.

2. AVDBANew commenter

Restatement:

A global leader of Educational Services has decided to form, worldwide, 15 commissions
to improve these services. N specialists were chosen to compose the commissions. Each
commission has the same number of members. Each specialist must belong to two
commissions and each pair of commissions must have, exactly, one member in common.

Determine: a) N, the number of specialists; b) the number of specialists within
each commission.

3. AVDBANew commenter

I would like to propose the following hints:

* The number of specialists is the summation over the number of commissions, excluding common members.

4. AVDBANew commenter

Having in mind a problem for working with progressions, I would like to suggest the following solution.

Consider a commission C. Let n be the number of members within this commission. A member within C must belong to C and to another commission, since each member must to belong to 2 commissions. This must occur to each member of C. Excluding C, there exist 14 commissions, so C must have n = 14 members. To see the pattern, consider a case, e.g., having 3 commissions. For this case, n would be 2, since:

Commission 1: Members: A, B
Commission 2: Members: A, ?
Commission 3: Members: B, ?

The ? above cannot be A, since A is belonging to 2 commissions as required, and cannot be B, since B is also belonging to 2 commissions as required. So, one reaches:

Commission 1: Members: A, B: n - 0 new members = 2 new members (A and B)
Commission 2: Members: A, D: n - 1 new members = 1 new members (D)

...

Commission 3: Members: B, D: 0 new members = 0 new members (No new member)

Adding the quantity of new members, 2 + 1 + 0 = 3, gives the total number N of members for this example. The pattern for our case follows the same reasoning, giving the progression, the summation over the 15 commisions excluding previously paired members (considering new
members):

(14 - 0) + (14 - 1) + (14 - 2) + ... + (14 - 14) = N

Since this is an Arithmetic Progression having 15 terms, being the first term 14 and the last one 0, we reach:

N = [(14 + 0) x 15] / 2 = 7 x 15 = 105.