At the detailed level we have microstates composed of a set of N items that are

Question

At the detailed level we have microstates composed of a set of N items that are in one of two conditions. We can name the two conditions heads and tails using coins as an analogy.
At the summary level we can characterize the macrostate of the system by the number of heads (independent of which coins are heads). The possible number of heads ranging from 0 to N.
We have N+1 macrostates (that don't specify the condition of each, individual coin) and

2N

microstates where the condition of each, individual coin is specified.
(A) When we have 22 items. What is the total number of microstates?  
(B) With 22 items how many microstates have all heads?  

(C) With 22 items, how many microstates have 11 heads?  

(D) The macrostates that are near equal numbers of heads and tails are the most likely. What fraction of the total number of microstates with 22 items are in one of these three macrostates: 10 heads, 11 heads, or 12 heads?
fraction of the microstates are in this 0.136 fraction of the macrostates.

Some nCr values that might be useful:
22 nCr 6 = 74,613            22 nCr 7 = 170,544
22 nCr 8 = 319,770            22 nCr 9 = 497,420
22 nCr 10 = 646,646            22 nCr 11 = 705,432
22 nCr 12 = 646,646            22 nCr 13 = 497,420
22 nCr 14 = 319,770            22 nCr 15 = 170,544
22 nCr 15 = 319,770            22 nCr 16 = 170,544

Explanation / Answer

A-:)

22 items that means asuming 22 coins ,N=22 possible total no. of microstates be T= 2^N=2^22 =4194304 states

B:-)

Every microstate can be head or tail. 22 heads from 22 coins=22C22= 1

=> only one way possible for all coins to be in head.

C:-)

Out of 22 any 11 coins be chosen and set them to head and remaining to be tails.

Choosing any 11 from 22 is nothing but 22C11 =705,432 ways

D:-)

10 heads= 22C10, 11 heads=22C11, 12 heads= 22C12

=> let these states be S=646,646+705,432+646,646=1998724, T=2^22 (from question A)

fraction = S/T = 1998724/4194304= 0.4765

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