n C 10-4 n C 9-6 n C 8-4 n C 7- n C 6-A- n-6 C10- n-2 C9- n-4 C10D- n-8 C8

The correct option is C We know, ^nC_r=^nC_n r ^nC_r+^nC_r 1=n+1^C_r ^nC_10+4^nC_9+6^nC_8+4^nC_7+^nC_6 4^nC_9 can be written as ^nC_7+3^nC_7 Similarly, ^4C_7 ...

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Byju's Answer

Standard XII

Mathematics

Binomial Coefficients

n C 10+4 n C ...

Question

$^{n} C_{10} + 4^{n} C_{9} + 6^{n} C_{8} + 4^{n} C_{7} +^{n} C_{6} =$

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Solution

The correct option is C

We know,
$^{n} C_{r} =^{n} C_{n - r}$
$^{n} C_{r} +^{n} C_{r - 1} = {n + 1}_{r}^{C}$
$^{n} C_{10} + 4^{n} C_{9} + 6^{n} C_{8} + 4^{n} C_{7} +^{n} C_{6}$
$4^{n} C_{9}$ can be written as $^{n} C_{7} + 3^{n} C_{7}$
Similarly, $^{4} C_{7}$ can be written as $^{n} C_{7} + 3^{n} C_{7}$
$=^{n} C_{10} +^{n} C_{9} + 3^{n} C_{9} + 6^{n} C_{8} + 3^{n} C_{7} +^{n} C_{7} +^{n} C_{6}$
write $^{n} C_{10} +^{n} C_{9} =^{n + 1} C_{9} a n d^{n} C_{7} +^{n} C_{6} =^{n + 1} C_{6}$
$=^{n} C_{10} +^{n} C_{9} + 3^{n} C_{9} + 6^{n} C_{8} + 3^{n} C_{7} +^{n} C_{7} +^{n} C_{6}$
$=^{n + 1} C_{10} +^{n} C_{9} + 3 (^{n} C_{9} + 2^{n} C_{8} +^{n} C_{7}) +^{n + 1} C_{7}$
$=^{n + 1} C_{10} + 3 (^{n} C_{9} +^{n} C_{8} +^{n} C_{8} +^{n} C_{7}) +^{n + 1} C_{7}$
$=^{n + 1} C_{10} + 3^{n + 1} C_{9} + 2^{n + 1} C_{9} + 2^{n + 1} C_{8} +^{n + 1} C_{8} +^{n + 1} C_{7}$
$=^{n + 2} C_{10} + 2^{n + 2} C_{9} +^{n + 2} C_{8}$
$=^{n + 2} C_{10} +^{n + 2} C_{9} +^{n + 2} C_{9} +^{n + 2} C_{8}$
$=^{n + 3} C_{10} +^{n + 3} C_{9}$
$=^{n + 4} C_{10}$

Suggest Corrections

nC10+4nC9+6nC8+4nC7+nC6=

$^{n} C_{10} + 4^{n} C_{9} + 6^{n} C_{8} + 4^{n} C_{7} +^{n} C_{6} =$