\[\frac{_{r-1}^{2n-1}\textrm{C}}{_{r}^{2n}\textrm{C}}=\left (\frac{2 n!}{ \left (r-1 \right )! \left (2 n -r + 1 \right )!} \right ) \left (\frac{r! \left (2 n - r \right )!}{2 n !} \right )=\frac{r}{2n - r + 1} \]
Boundary condition \[1\leq r\leq n \]
\[\frac{r}{2 n - r + 1} < 1 \]
In general \[_{r}^{n}\textrm{C} \] represents combination of n things taken r at a time.
In other words selection of r things out of n things
For example \[_{2}^{5}\textrm{C} \] means how many outcombination of 5 things is possible taken 2 objects at a time
Examples-
1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Explanation:
Number of ways of selecting 3 consonants from 7
= 7C3
Number of ways of selecting 2 vowels from 4
= 4C2
Number of ways of selecting 3 consonants from 7 and 2 vowels from 4
= 7C3 × 4C2
=(7×6×53×2×1)×(4×32×1)=210=(7×6×53×2×1)×(4×32×1)=210
It means we can have 210 groups where each group contains total 5 letters (3 consonants and 2 vowels).
Example 2.
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Hence we have 4 options as given below
We can select 4 boys ...(option 1)
Number of ways to this = 6C4
We can select 3 boys and 1 girl ...(option 2)
Number of ways to this = 6C3 × 4C1
We can select 2 boys and 2 girls ...(option 3)
Number of ways to this = 6C2 × 4C2
We can select 1 boy and 3 girls ...(option 4)
Number of ways to this = 6C1 × 4C3
Total number of ways
= 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3
= 6C2 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C1[∵ nCr = nC(n-r)]
=6×52×1+6×5×43×2×1×4=6×52×1+6×5×43×2×1×4 +6×52×1×4×32×1+6×4+6×52×1×4×32×1+6×4
=15+80+90+24=209
