If we additionally fix the first letter, we count only all combinations containing element ’1’, which must then be{ <hi xmlns="http://www.tei-c.org/ns/1.0" rend="italic">w-1</hi> /choose <hi xmlns="http://www.tei-c.org/ns/1.0" rend="italic">a-1</hi> <hi xmlns="http://www.tei-c.org/ns/1.0" rend="italic">}</hi>, since we have fixed the first letter and from the remaining <hi xmlns="http://www.tei-c.org/ns/1.0" rend="italic">w - 1 </hi>letters, we can choose any <hi xmlns="http://www.tei-c.org/ns/1.0" rend="italic">a - 1</hi> remaining elements of the abbreviation. This restricts the possible numbers already considerably. Overall, we halve possibilities, so 2^{w-1} - 1 becomes the sum formula, which would still leave us with 16,383 possibilities for a word of 15 letters.