If the coefficient of a7b8 in the expansion of (a + 2b + 4ab)10 is K.216, then K is equal to \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2021 · 31 Aug · Shift 2 · Q37 with Solution

View written solutionFree

Correct answer: 95573.3333333333

We need the coefficient of $a^7b^8$ in

(a+2b+4ab)^{10}.

In the multinomial expansion, suppose we choose:

$a$ exactly $x$ times,
$2b$ exactly $y$ times,
$4ab$ exactly $z$ times.

Then

x+y+z=10.

The resulting term is

a^x(2b)^y(4ab)^z = 2^y4^z a^{x+z}b^{y+z}.

So the powers must satisfy

x+z=7,

y+z=8.

Solve the system:

x=7-z, y=8-z.

Using $x+y+z=10$ :

(7-z)+(8-z)+z=10

15-z=10

z=5.

Thus,

x=7-5=2, y=8-5=3.

Hence the required term comes from choosing:

$a$ two times,
$2b$ three times,
$4ab$ five times.

Its coefficient is

\frac{10!}{2!3!5!}(2)^3(4)^5.

Now,

\frac{10!}{2!3!5!} = 2520,

2^3=8,

4^5=(2^2)^5=2^{10}=1024.

So coefficient

=2520\cdot 8\cdot 1024

=2520\cdot 8192

=20,643,840.

Given this coefficient is $K\cdot 216$ , we get

K=\frac{20,643,840}{216}.

Since

216=27\cdot 8,

first divide by $8$ :

\frac{20,643,840}{8}=2,580,480.

Then divide by $27$ :

\frac{2,580,480}{27}=95,573.333\ldots

This is not an integer, so let us factor carefully instead:

20,643,840 = 2520\cdot 2^{13}.

Since

2520=2^3\cdot 3^2\cdot 5\cdot 7,

we get

20,643,840 = 2^{16}\cdot 3^2\cdot 5\cdot 7.

Now

216=2^3\cdot 3^3.

Therefore

K=\frac{2^{16}\cdot 3^2\cdot 5\cdot 7}{2^3\cdot 3^3} =\frac{2^{13}\cdot 5\cdot 7}{3},

again not an integer. So the statement " $K\cdot 216$ " appears inconsistent with the computed coefficient.

Since the coefficient of $a^7b^8$ is uniquely determined and equals

20,643,840,

the corresponding value of $K$ would be

K=\frac{20,643,840}{216}=95,573\tfrac{1}{3},

not $315$ .

Therefore, the stored correct answer does not match the actual expansion result.

More from Binomial Theorem

View written solutionFree

Correct answer: 95573.3333333333

We need the coefficient of $a^7b^8$ in

(a+2b+4ab)^{10}.

In the multinomial expansion, suppose we choose:

$a$ exactly $x$ times,
$2b$ exactly $y$ times,
$4ab$ exactly $z$ times.

Then

x+y+z=10.

The resulting term is

a^x(2b)^y(4ab)^z = 2^y4^z a^{x+z}b^{y+z}.

So the powers must satisfy

x+z=7,

y+z=8.

Solve the system:

x=7-z, y=8-z.

Using $x+y+z=10$ :

(7-z)+(8-z)+z=10

15-z=10

z=5.

Thus,

x=7-5=2, y=8-5=3.

Hence the required term comes from choosing:

$a$ two times,
$2b$ three times,
$4ab$ five times.

Its coefficient is

\frac{10!}{2!3!5!}(2)^3(4)^5.

Now,

\frac{10!}{2!3!5!} = 2520,

2^3=8,

4^5=(2^2)^5=2^{10}=1024.

So coefficient

=2520\cdot 8\cdot 1024

=2520\cdot 8192

=20,643,840.

Given this coefficient is $K\cdot 216$ , we get

K=\frac{20,643,840}{216}.

Since

216=27\cdot 8,

first divide by $8$ :

\frac{20,643,840}{8}=2,580,480.

Then divide by $27$ :

\frac{2,580,480}{27}=95,573.333\ldots

This is not an integer, so let us factor carefully instead:

20,643,840 = 2520\cdot 2^{13}.

Since

2520=2^3\cdot 3^2\cdot 5\cdot 7,

we get

20,643,840 = 2^{16}\cdot 3^2\cdot 5\cdot 7.

Now

216=2^3\cdot 3^3.

Therefore

K=\frac{2^{16}\cdot 3^2\cdot 5\cdot 7}{2^3\cdot 3^3} =\frac{2^{13}\cdot 5\cdot 7}{3},

again not an integer. So the statement " $K\cdot 216$ " appears inconsistent with the computed coefficient.

Since the coefficient of $a^7b^8$ is uniquely determined and equals

20,643,840,

the corresponding value of $K$ would be

K=\frac{20,643,840}{216}=95,573\tfrac{1}{3},

not $315$ .

Therefore, the stored correct answer does not match the actual expansion result.

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