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Mathematics Question on permutations and combinations

If nCr−1=28^{n}C_{r-1}=28, nCr=56^{n}C_{r}=56 and nCr+1=70^{n}C_{r+1}=70, then the value of rr is equal to

A

11

B

22

C

33

D

44

Answer

33

Explanation

Solution

nCr−1=28,nCr=56⋅nCr+1=70{ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56 \cdot{ }^{n} C_{r+1}=70
⇒ nCrnCr−1=5628\Rightarrow\, \frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{56}{28}
⇒ n!r(r−1)!(n−r)!n!(r−1)!(n−r+1)(n−r)!=n−r+1=2r\Rightarrow \, \frac{\frac{n !}{r(r-1) !(n-r) !}}{\frac{n !}{(r-1) !(n-r+1)(n-r) !}}{=n-r+1=2 r}
⇒ n−3r=−1     …(i)\Rightarrow\, n-3 r=-1\,\,\,\,\,\dots(i)
nCr+1nCr=7056\frac{{ }^{n} C_{r+1}}{{ }^{n} C_{r}}=\frac{70}{56}
⇒ n!(r+1)!(n−r−1)!n!n!r!(n−r)!=3528\Rightarrow\, \frac{n !}{\frac{(r+1) !(n-r-1) !}{n !}}{\frac{n !}{r !(n-r) !}}=\frac{35}{28}
⇒ (r+1)r!(n−r−1)!n!35r!(n−r)(n−r−1)!=3528\Rightarrow\, \frac{\frac{(r+1) r !(n-r-1) !}{n !}}{\frac{35}{r !(n-r)(n-r-1) !}}=\frac{35}{28}
⇒ n−rr+1=3528\Rightarrow\,\frac{n-r}{r+1}=\frac{35}{28}
⇒ 28n−28r=35r+35\Rightarrow \, 28 n-28 r=35 r+35
⇒ 28n−63r=35      …(ii)\Rightarrow\, 28 n-63 r=35\,\,\,\,\,\,\dots(ii)
Multiply by 21 in E (i) and subtracting from E (ii),
21n−63r=−2121 n-63 r=-21
28n−63r=3528 n-63 r=35
−+−−7n=−56⇒n=8\frac{-\quad+\quad-}{-7 n=-56 \Rightarrow n=8}
From E (i), 3r=n+1=8+13 r=n+1=8+1
r=3r=3