Combinatoric in Accounting

Prove:
\displaystyle\sum_{n=2}^{n}{_n}C_r = 2^n-1-n
\displaystyle\sum_{n=2}^{n}{_n}C_r 2^r= 3^n-1-2n
Note:
{_4}C_2=\frac {4.3}{2!}
Proof:
1.
\displaystyle\sum_{n=2}^{n}{_n}C_r =(1+1)ⁿ -1- \displaystyle {_n}C_1 =2^n-1-n
2.
\displaystyle\sum_{n=2}^{n}{_n}C_r 2^{r}=(1+2)ⁿ -1 – \displaystyle {_n}C_1 .2^{1}= 3^n-1-2n

An accounting transaction = Debit p accounts + Credit q accounts (p, q ≥ 1)

In a company with total n accounts,
Prove: there are 3^n -2^{n+1} + 2 transactions.

Proof:
Let the Set of all accounts T = {a1, a2, …, an}
aj = account with value ‘+’ (credit), or ‘-‘ (debit), or 0 (nil)
1. Trivial transaction: To= {0, 0,…..0} = 1 way
2. Choose r accounts from n = \displaystyle {_n}C_r
3. Slot ‘+’ or ‘-‘ in these r accounts = 2^r ways
exclude 2 impossible all ‘+’, ‘-‘ transactions = 2^r- 2 ways
4.
Let T1 = \displaystyle\sum_{n=2}^{n}{_n}C_r .(2^r-2)
Total transaction = To + T1
= \displaystyle 1+ \sum_{n=2}^{n}{_n}C_r .(2^r-2)
{Apply previous results}
= \displaystyle 1+ \sum_{n=2}^{n}{_n}C_r .2^r -2.\sum_{n=2}^{n}{_n}C_r
= 1+3^n-1-2n -2(2^n-1-n)
= 3^n -2^{n+1} + 2  [QED]

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