# Selections math problems

1. ### Wiz

21
I am stuck on this prob.....16 men are seated along a round table....in how many ways can i select 7 out of em such tht none of the selected individual were sitting next to each other?...
thanks

2. ### Timbuqtu

83
In how many ways could you select 8 men? In how many fundamentally different ways (that is, without being able to rotate/flip one into another)can you select 7 men?

3. ### Wiz

21
i dont get u.....wht are u trying to say??.are u challenging the question or asking me a question which i myself posted on the forum?...what??

4. ### Timbuqtu

83
Hmmm...I was trying to be pedagogical. See if I could point you in the right direction by asking some slightly different questions, rather than just stating the solution.

Another try: We have to select 7 persons with gaps (of at least one person) between them. This is equivalent to selecting 7 gaps (of at least one person) with exaclty one person in between. Now, in how many ways can you distribute the remaining 9 persons over the 7 gaps? We've got two different cases:

(1) We've got a gap of three persons, like: #000#0#0#0#0#0#0 (where # denotes selected). In how many ways can you do this?

(2) We've got two gaps of two persons, like: #00#00#0#0#0#0#0. In how many ways can you do this?

5. ### Timbuqtu

83
By the way: I get 14 + 42 = 56 possible ways.

6. ### Wiz

21
hmm...i get ur method..but ur ans is wrong..check it again..in the meanwhile i'll give it another shot..do post ur corrected ans...thanks.

7. ### Timbuqtu

83
Oh yes, I see, I made a mistake.

(1) In this case the selection is characterized by the position (of the middle person) of the gap of three. So there are 16 different possible selections.

(2) Now the selection is characterized by the position of the two gaps of two persons. The 'first' gap (of two) can be chosen arbitrary, so 16 possibilities. The second gap (of two) can be chosen from the 7-1=6 remaining gaps between selected persons. Now we have to devide by two because the gaps (of two) could interchanged to yield the same selection. So we get 16 x 6 / 2 = 48 possible selections.

Conclusion there are 64 possible selections.

8. ### Wiz

21
ok thtz correct.....but is there a way to generalise it??...say selection of r persons out of n ppl??.(in the same prob)

9. ### VietDao29

1,422
First, build a matrix like
$$A = \left[ \begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} & . . . \\ a_{21} & a_{22} & a_{23} & a_{24} & . . . \\ a_{31} & a_{32} & a_{33} & a_{34} & . . . \\ \vdots & \vdots & \vdots & \vdots & . . . \end{array} \right]$$
Where $$a_{1j} = 1$$ and $$a_{n1} = 1$$, and define : $$a_{ij} = a_{(i - 1)j} + a_{i(j - 1)}$$
$$A = \left[ \begin{array}{ccccc} 1 & 1 & 1 & 1 & . . . \\ 1 & 2 & 3 & 4 & . . . \\ 1 & 3 & 6 & 10 & . . . \\ \vdots & \vdots & \vdots & \vdots & . . . \end{array} \right]$$
Problem: n people in a round table. No 2 selected people sits next to each other. Select p people.
Calculate: $$\delta = n - (2(p - 1) + 1)$$. Then the total number of ways is :
$$C = a_{\delta p} + \sum_{x = 1}^{\delta} a_{xp}$$
I wonder if anyone has discovered this. Anyway, if noone else has. Let’s call it ‘The little theorem of Viet Dao’.
EDIT : This formula can be horribly wrong if there is no way to select.
Viet Dao,

Last edited: Aug 2, 2005
10. ### Wiz

21
i understood absolutely nothing in ur previous post......could u plz explain??....

11. ### VietDao29

1,422
Okay, a matrix m x n is defined as:
$$A_{m x n} = \left[ \begin{array}{ccccc} a_{11} & a_{12} & a_{13} & . . . & a_{1n} \\ a_{21} & a_{22} & a_{23} & . . . & a_{2n} \\ \vdots & \vdots & \vdots & \vdots & . . . \\ a_{m1} & a_{m2} & a_{m3} & . . . & a_{mn} \end{array} \right]$$
$$a_{ij}$$ 0 < i < m, 0 < j < n is one item in the matrix. For example, i = 1, j = 1, then you will have $$a_{ij} = a_{11}$$
And my matrix will have $$a_{1j} = 1 \mbox{ and } a_{i1} = 1$$
So it will look like:
An item $$a_{ij}$$ is the one that lies on the i row, and the j column of the matrix.
$$A = \left[ \begin{array}{ccccc} 1 & 1 & 1 & 1 & . . . \\ 1 & a_{22} & a_{23} & a_{24} & . . . \\ 1 & a_{32} & a_{33} & a_{34} & . . . \\ \vdots & \vdots & \vdots & \vdots & . . . \end{array} \right]$$
Then I say : $$a_{ij} = a_{(i - 1)j} + a_{i(j - 1)}$$
Example : $$a_{22} = a_{12} + a_{21} = 1 + 1 = 2$$
$$a_{23} = a_{13} + a_{22} = 1 + 2 = 3$$
The item $$a_{ij}$$ is the sum of the item above it and the item to the left of it.
And so on, so my matrix will look like:
$$A = \left[ \begin{array}{ccccc} 1 & 1 & 1 & 1 & . . . \\ 1 & 2 & 3 & 4 & . . . \\ 1 & 3 & 6 & 10 & . . . \\ \vdots & \vdots & \vdots & \vdots & . . . \end{array} \right]$$
Then I calculate : $$\delta = n - (2(p - 1) + 1)$$. There are 2 cases:
1. $$\delta \leq 0$$ : No selection.
2. $$\delta > 0$$:
Then the number of selection will be:
$$C = a_{\delta p} + \sum_{x = 1}^{\delta} a_{xp}$$
It means:
$$C = a_{\delta p} + a_{1p} + a_{2p} + a_{3p} + ... + a_{\delta p}$$
Where p is the number of people you want to select, n is the total number of people sitting on the round table.
* Solving your problem using my method:
$$\delta = n - (2(p - 1) + 1) = 3$$
p = 7, n = 16 (This is what you have total people = 16, number of selected people = 7).
$$A = \left[ \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1. . . \\ 1 & 2 & 3 & 4 & 5 & 6 & 7. . . \\ 1 & 3 & 6 & 10 & 15 & 21 & 28. . . \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & . . . \end{array} \right]$$
So you will have:
$$C = a_{\delta p} + a_{1p} + a_{2p} + a_{3p} + ... + a_{\delta p}$$
$$= a_{37} + a_{17} + a_{27} + a_{37} = 28 + 28 + 7 + 1 = 64.$$
So there are 64 different ways to randomly choose 7 people, in which in every 2 selected people, they don't sit next to each other.
You can try another number of people sitting on the table (n), and the number of selected people (p).
Viet Dao,

Last edited: Aug 3, 2005