Q.
A box contains tickets numbered from 0 to 999999 . The probability of getting a ticket- with sum of digits on it being equal to 21 is given by 106PC5−6⋅qC5+15⋅rC5, find the value of (p+q+r).
Concept: Coefficient if xr in (1−x)−n=n+r−1Cr n(A)= coefficient of x21 in (1+x+x2+……..+x9)6
(six digit starting with 000000,000001 etc. each place can have digit from 0 to 9) ∴n(A)= coefficient of x21 in (1−x1−x10)6 n(A)= coefficient of x21 in (1−6C1x10+6C2x20)(1−x)−6 = coefficient of x21 in (1−6x10+15x20)(1−x)−6 = coefficient of x21 in (1−x)−6−6⋅ coefficient of x11 in (1−x)6+15 coefficient of x in (1−x)−6 =26C21−6⋅16C11+15⋅6C1 =26C5−6⋅16C5+15⋅6C5 ⇒p+q+r=26+16+6=48