Name: Counting With Bijections
Uploaded: 2019-07-19
Duration: 11 min 43 s
Description: In this informative video, you will learn about the Bijection principle in algebra. Let's watch, understand and master this concept right away!

Question

For which positive integral values of

n

can the set

\{1, 2, 3, \dots, 4 n\}

be split into

n

disjoint

4

-element subsets

\{a, b, c, d\}

such that in each of these sets

a = \frac{(b + c + d)}{3}

.

Accepted Answer

Suppose $\{a, b, c, d\}$ is a group in which $a = \frac{(b + c + d)}{3}$ .

$\Rightarrow b + c + d = 3 a$

$\Rightarrow a + b + c + d = 4 a . . . . (i)$

$\Rightarrow$ Sum of four elements of any subset is a multiple of $4$ .

As the set $\{1, 2, 3, \dots, 4 n\}$ will be partitioned into $n$ disjoint $4 -$ element subsets, so the property is applicable on the elements of these disjoint subsets too.

So, in case if such an $n$ exists, then $1 + 2 + . . . + 4 n$ is divisible by $4$ .

Here, using sum of an arithmetic progression, $1 + 2 + . . . + 4 n = (\frac{4 n}{2}) (4 n + 1) = 2 n (4 n + 1)$ , which is an even number.

Thus a necessary condition for existence of such a set is that $n$ be even.

Now let's try to prove that is sufficient also.

Let $n = 2 k$ for some $k$ , then set becomes $\{1, 2, 3, . . ., 8 k\}$ .

Now let's partition the set $\{1, 2, 3, . . ., 8 k\}$ into $k$ groups or disjoint subsets of $8$ consecutive integers each.

Any group will look like this $\{a + 1, a + 2, a + 3, . . ., a + 8\}$ , further divide this group into two sub-groups containing four elements each as following:

$\{a + 4, a + 1, a + 3, a + 8\}$ , $\{a + 5, a + 2, a + 6, a + 7\}$ .

For both groups it can be verified that $(a + 4) + (a + 1) + (a + 3) + (a + 8) = 4 (a + 4) & (a + 5) + (a + 2) + (a + 6) + (a + 7) = 4 (a + 5)$ , which are satisfying the condition $(i)$ .

Hence, for $n$ being positive even integer desired partition is always possible.

For which positive integral values of $n$ can the set $\{1, 2, 3, \dots, 4 n\}$ be split into $n$ disjoint $4$ -element subsets $\{a, b, c, d\}$ such that in each of these sets $a = \frac{(b + c + d)}{3}$ .

Important Questions on Combinatorics

Learn and Prepare for any exam you want !

For which positive integral values of n can the set 1,2,3,…,4n be split into n disjoint 4-element subsets a,b,c,d such that in each of these sets a=b+c+d3.

Important Questions on Combinatorics

Learn and Prepare for any exam you want !

For which positive integral values of $n$ can the set $\{1, 2, 3, \dots, 4 n\}$ be split into $n$ disjoint $4$ -element subsets $\{a, b, c, d\}$ such that in each of these sets $a = \frac{(b + c + d)}{3}$ .