I do understand the proof of C(n,r)=C(n,n-r) but cannot catch the intuition behind this.Intuitively why would the number of ways to select r objects from a set of n objects be equal to the number of ways to select (n-r) objects from a set of n objects?
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When you choose $r$ from $n$ objects, you automatically choose the $n-r$ that are left out.
More precisely, if $U$ is a set of $n$ elements and $A(k)$ is the set of subsets of size $k$ , then $X \mapsto U \setminus X$ defines a bijection $A(k) \to A(n-k)$.
lhf
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i could not understand how n-r objects are chosen automatically. – MrAP Nov 24 '16 at 17:03
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@MrAP, taking the complementary. – Martín-Blas Pérez Pinilla Nov 24 '16 at 18:34
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1@MrAP Choosing $n$ objects to form a set is the same as choosing $n-r$ to omit from that set. – amd Nov 24 '16 at 22:39
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@Martín-BlasPérezPinilla, i understood that n-r objects are chosen automatically but why are the number of combinations in both case equal? – MrAP Nov 25 '16 at 23:03
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Let's list all the subsets of size $2$ of the set $\{a,b,c,d,e\}$, and also all the subsets of size $3$:
\begin{align} \text{Size $2$} & \qquad \qquad \quad \text{Size $3$} \\ \{a,b\} & \quad \longleftrightarrow\quad \{ c,d,e\} \\ \{a,c \} & \quad \longleftrightarrow\quad \{b,d,e\} \\ \{a,d\} & \quad \longleftrightarrow\quad \{b,c,e \} \\ \{a,e \} & \quad \longleftrightarrow\quad \{b,c,d\} \\ \{b,c\} & \quad \longleftrightarrow\quad \{a,d,e\} \\ \{b,d\} & \quad \longleftrightarrow\quad \{a,c,e\} \\ \{b,e\} & \quad \longleftrightarrow\quad \{a,c,d\}\\ \{c,d\} & \quad \longleftrightarrow\quad \{a,b,e\} \\ \{c,e \} & \quad \longleftrightarrow\quad \{a,b,d\} \\ \{d,e\} & \quad \longleftrightarrow\quad \{a,b,c\} \end{align}
We see very clearly that there is a perfect 1-1 correspondence between the subsets of size $2$, and the subsets of size $3$. Every subset of size $2$ gets paired up with its complementary set.
littleO
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Suppose, there are five balls in a box numbered 1 to 5. A person wants to distribute two balls to two childs who reside at Jadavpur and three balls to another childs who reside at Golpark. Let he gives away two balls numbered 1 & 2 to two childs who reside at Jadavpur and he gives away three balls numbered 3,4 & 5 to two childs who reside at Golpark. he may give away two balls numbered 1 & 3 to two childs who reside at Jadavpur and he gives away three balls numbered 2,4 & 5 to two childs who reside at Golpark. So, whenever he choose any two balls, automatically three balls are choosen. Therefore, the number of ways selecting two balls is equal to the number of ways selecting three balls.
So, C(5,2)=C(5,3) or, C(5,2)=C(5,5-2)
If 5 is replased by n and 5 is replased by r. It may be generalised that, C(n,r)=C(n,n-r) .
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Just imagine you have a basket with 10 different pieces of fruit. For every way you have to choose 3 of them to eat, there are 7 left in the basket.
J. C.
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