The set X has (2 m + 1) elements. The number of subsets of X having atleast one element and at the most 'm' elements is 16383. Find the value of m.

Number of ways in which one element can be selected = $$^{2m+1}C_1$$
Number of ways in which two elements can be selected = $$^{2m+1}C_2$$
...
Number of ways in which 'm' elements can be selected = $$^{2m+1}C_m$$
So, total number of ways in which the subsets can be formed = $$^{2m+1}C_1$$ + $$^{2m+2}C_1$$ +...+ $$^{2m+1}C_m$$
We have $$^{2m+1}C_0$$ + $$^{2m+1}C_1$$ + $$^{2m+1}C_2$$ + ... + $$^{2m+1}C_{2m+1}$$ = $$2^{2m+1}$$

Also, $$^{2m+1}C_{m+1} = ^{2m+1}C_m$$, $$^{2m+1}C_{m+2} = ^{2m+1}C_{m-1}$$...$$^{2m+1}C_{2m+1} = ^{2m+1}C_0$$
So, 2*$$(^{2m+1}C_0 + ^{2m+1}C_1 + ... + ^{2m+1}C_m) = 2^{2m+1}$$
=> $$^{2m+1}C_0 + ^{2m+1}C_1 + ^{2m+1}C_2 + .... + ^{2m+1}C_m = 2^{2m+1}/2$$
=> $$^{2m+1}C_1 + ^{2m+1}C_2 + .... + ^{2m+1}C_m = 2^{2m+1}/2 - ^{2m+1}C_0 = 2^{2m+1}/2 - 1$$
=> $$2^{2m+1}/2 - 1$$ = 16383
=> $$2^{2m+1}/2$$ = 16384
=> $$2^{2m+1} = 32768 = 2^{15}$$
=> (2m+1) = 15
=> m = 7

Now understood..
A nice way to do that. thank you sir