How to  calculate these summations? How to find the values of these kind of summations:
$$\large\sum_{i=0}^6(6-i)\;\ast\;\sum_{j=1}^6(7-j)\;\ast\;\sum_{k=2}^7(8-k)\;\ast\;\sum_{\ell=3}^8(9-\ell)$$
 A: Hint: Use the following formula.


*

*$\displaystyle \sum\limits_{i=1}^{n} i = \frac{n \cdot (n+1)}{2}$.

*$\displaystyle \sum\limits_{i=0}^{6} (6-i) = \sum\limits_{i=0}^{6} (6 - i) = 6 \cdot \bigl(1+1+1+1+1+1+1\bigr)- \sum_{i=0}^{6} i =6 \times 7 - \frac{6 \cdot 7}{2}$
Similarly you can calculate the other summation's  as well and then multiply them. 
A: Or the formula:
$$\sum_{i=k_1}^{k_2} (p-i) =  (k_2-k_1+1)(p-\frac{1}{2}(k_1+k_2)).$$
A: Use that $$\begin{align}\sum_{t=a}^b(c-t)&=\left(\sum_{t=a}^bc\right)-\left(\sum_{t=a}^b t\right)\\\\&=\left(\sum_{t=a}^bc\right)-\left(\sum_{s=0}^{b-a} (s+a)\right)\\\\ &=\left(\sum_{t=a}^bc\right)-\left(\sum_{s=0}^{b-a} s\right)-\left(\sum_{s=0}^{b-a}a\right)\\\\ &=(b-a+1)c-(b-a+1)a-\left(\sum_{s=0}^{b-a} s\right)\\\\&=(b-a+1)(c-a)-\left(\sum_{s=0}^{b-a} s\right)\\\\&=(b-a+1)(c-a)-\frac{(b-a)(b-a+1)}{2}\\\\&=(b-a+1)(c-a-\tfrac{b}{2}+\tfrac{a}{2})\\\\&=(b-a+1)(c-\tfrac{b}{2}-\tfrac{a}{2})\end{align}$$
for each term. For example,
$$\sum_{i=0}^6(6-i)=6+5+4+3+2+1=\fbox{21}=7\cdot 3=(6-0+1)(6-\tfrac{6}{2}-\tfrac{0}{2})\qquad\checkmark$$
