This makes the multiplicity when the dye is introduced (t = 0) the total number of ways of placing five dye molecules in nine different spaces.
Similar to arranging books by subject, the total number of ways is given by 10! or 3,628,800 because any book can be first, second, third, and so on.
The total number of ways that g common auxiliary nodes randomly select initial keys is, where for is defined in (6).
Hence, the total number of ways by which V randomly selects root keys from the root key pool is given by.
Hence, the total number of ways that the common auxiliary node selects initial keys from initial key pool should be (6).
When two neighboring ordinary nodes U and V have g common auxiliary nodes, the total number of ways that U and V share m derived keys can be calculated as follows: First, U and V have different ways to randomly select root keys from root key pool.
Thus, considering the same data block of size R × Q × N bits and the same memory M for all three (MBTC, NBTC, and MNBTC), in Table 1, we present the total number of ways from component encoders' trellises (MBCE, NBCE, and MNBCE).
Thus, the total number of ways to put a ball in an empty place on the Lth transition is m 1 × (m - 1 ) L - 1.
The total number of ways of partitioning a set of n distinct elements into non-empty disjoint (and unlabelled) subsets, is given by the sum B (n ) = ∑ k = 1 n { n k }.
But we altered the amount of money by which someone might be better or worse off in a total of 11 ways.
