Solve the following

Question:

If n +2C8 : n − 2P4 = 57 : 16, find n.

Solution:

We have, ${ }^{n+2} C_{8}:{ }^{n-2} P_{4}=57: 16$

$\Rightarrow \frac{{ }^{n+2} C_{8}}{{ }^{n-2} P_{4}}=\frac{57}{16}$

$\Rightarrow \frac{(n+2) !}{8 !(n-6) !} \times \frac{(n-6) !}{(n-2) !}=\frac{57}{16}$

$\Rightarrow \frac{(n+2)(n+1) n(n-1)(n-2) !}{8 !} \times \frac{1}{(n-2) !}=\frac{57}{16}$

$\Rightarrow(n+2)(n+1) n(n-1)=\frac{57}{16} \times 8 !=\frac{19 \times 3}{16} \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$\Rightarrow(n+2)(n+1) n(n-1)=143640$

$\Rightarrow(n-1) n(n+1)(n+2)=19 \times 3 \times 7 \times 6 \times 5 \times 4 \times 3$

$\Rightarrow(n-1) n(n+1)(n+2)=19 \times(3 \times 7) \times(6 \times 3) \times(4 \times 5)$

$\Rightarrow(n-1) n(n+1)(n+2)=18 \times 19 \times 20 \times 21$

$\Rightarrow n-1=18$

$\Rightarrow n=19$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now