Equal Distribution of Learners into Different Class Rooms

OGDEN's BASIC ENGLISH INSTITUTE

Equal Distribution of Learners into Different Class Rooms

Number of Learners in one class : 50
Four Different Class Rooms : Respectively 10, 10, 15, and 15 learners' rooms.
10, 10, 15, and 15 learners' rooms are equally divided for purpose of account into 5, 5, and 5 ; 5, 5, and 5 ; 5, 5, and 5 learners' rooms.
Each room, is given an alphabetical name in the alphabetical order as follows:- 5(a), 5(b), 5(c), 5(d) and 5(e), 5(f), 5(g), 5(h), 5(i), 5(j) This is because each class room is different when it is look at from the angle of distribution of learners into it. If these rooms are put together into what they were, they are as follows:- 5(a) + 5(b) = 10 = One class room 10 learners in it. 5(c) + 5(d) = 10 = One class room 10 learners in it. 5(e) + 5(f) + 5(g) = 15 = One class room 15 learners in it. 5(h) + 5(i) + 5(j) = 15 = One class room 15 learners in it.
When the room 5(a) takes every place of rooms 5(e), 5(f), 5(g), 5(h), 5(i), (5j) there are 7 ways of doing it as given below. This is the same with 5(b), 5(c), and 5(d) rooms. So a total number of combination of 5(a), 5(b), 5(c), 5(d) rooms with 5(e), 5(f), 5(g), 5(h), 5(i), 5(j) rooms is 7 x 4 = 28 ways and the case with 5(a) is as given below.
The combination is the way in which it is possible to make equal distribution of learners into 10, 10, 15, and 15 learners' rooms. Each of the learners in each room are given a number starting with 1 and ending with 50 so that learners are kept from getting mixed with each other between different rooms.
If English teaching is done at 56 times a year, these 28 ways are able to be used twice.

Number of Learners in one class : 50
Six Different Class Rooms : Respectively 10, 10, 10, 10, 5, and 5 learners' rooms.
10, 10, 10, 10, 5 and 5 learners' rooms are equally divided for purpose of account into 5, 5, 5, 5, 5, 5, 5, 5, and 5 learners' rooms.
Each room, is given an alphabetical name in the alphabetical order as follows:- 5(a), 5(b), 5(c), 5(d), 5(e), 5(f), 5(g), 5(h), and 5(i), 5(j)
This is the same reason as mentioned in the first example above 2.
If these rooms are put together into what they were, they are as follows: -
When the room 5(i) takes every place of rooms 5(a), 5(b), 5(c), 5(d), 5(e), 5(f), 5(g), 5(h), there are 9 ways of doing it as given below. This is the same with 5(j) room. So a total number of combination of 5(i), 5(j) rooms with 5(a), 5(b), 5(c), 5(d), 5(e), 5(f), 5(g), 5(h) rooms is 9 x 2 = 18 ways, as given below.
The combination is the way in which it is possible to make equal distribution of learners into 10, 10, 15, and 15 learners' rooms. Each of the learners in each room are given a number starting with 1 and ending with 50 so that learners are kept from getting mixed with each other between different rooms.
If English teaching is done at 51 times a year, these 17 ways are able to be used at three times.
Saburo Terada
Saitama Prefecture
Thursday, December 16, 2004
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The writing is for readers who find it hard to make such equal distribution after reading "Mass Individual Teaching".