Question

UCWA Decemt
If (n + 1) C +1:n C:(n-1) C-1 = 11:6:3
find the values of 'n' and​

Answer 1

sry I couldn't type it clearly

ans: n=2r

r=5

so n= 10

hope u get it

n+1

C

r+1

:

n

C

r

:

n−1

C

r−1

=11:6:3

Therefore,

n

C

r

n+1

C

r+1

=

6

11

∵

n

C

r

n+1

C

r+1

=

(r+1)!(n−r)!

(n+1)!

×

n!

r!(n−r)!

=

r+1

n+1

∴

r+1

n+1

=

6

11

.....(1)

Similarly,

n−1

C

r−1

n

C

r

=

r

n

=

3

6

n=2r.....(2)

Now from eq

n

(1)&(2), we have

r+1

2r+1

=

6

11

⇒6(2r+1)=11(r+1)

⇒12r+6=11r+11

⇒r=5