Show $ {}_{n}C_{4}={}_{n-1}C_{3}$ +${}_{n-2}C_{3} +{}_{n-3}C_{3} + \cdots + {}_{3}C_{3} $
where: ${}_{n}C_{i}$ is the number of ways of choosing $i$ elements from $n$
-I've been explicitly requested to provide a non algebraic proof; and to instead use a 'combinatorial argument.'
-'Combinatorial argument' refers to providing a synonymous situation that describes the above equation.
what I attempted:
-choosing $4$ balls from $n$ balls in an urn may be done in ${}_{n}C_{4}$ ways.
This is the same as:
1)st step: from n-1 balls choose 3
2)nd step: from n-2 remaining balls choose 3
.
.
.
n-3)th step: from 3 remaining balls choose 3
The events above are therefore the same as choosing 4 balls from n.
The above "proof" doesn't convince me if it is indeed proof.
Does anybody have insight into how this problem may be visualized using a 'combinatorial argument.'