If 100 C 50 can be prime factorised as 2- 3- 5 Y - 7- where - - y - - - are non-negative integers- then the CORRECT relations is are A- -B- Y - - y -1D- Y -0

The correct options are A α lt;β C α+δ=β+γ 1 D γ+δ=0^100C_50=(100)!(50!)(50!)=2^α· 3^β· 5^γ· 7^δ… where α, β, γ, δ, … are non negative integers. Exponent of ...

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Byju's Answer

Standard XII

Mathematics

Binomial Theorem

If 100 C 50 c...

Question

If $^{100} C_{50}$ can be prime factorised as $2^{α} \cdot 3^{β} \cdot 5^{γ} \cdot 7^{δ} \dots$ where $α, β, γ, δ, \dots$ are non-negative integers, then the CORRECT relation(s) is (are)

α < β

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γ < δ

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α + δ = β + γ - 1

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γ + δ = 0

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Solution

The correct options are
A $α < β$
C $α + δ = β + γ - 1$
D $γ + δ = 0$
$^{100} C_{50} = \frac{(100)!}{(50!) (50!)} = 2^{α} \cdot 3^{β} \cdot 5^{γ} \cdot 7^{δ} \dots$ where $α, β, γ, δ, \dots$ are non- negative integers.
Exponent of $2$ in $(100)!$ is
$[\frac{100}{2}] + [\frac{100}{2^{2}}] + [\frac{100}{2^{3}}] + \dots + [\frac{100}{2^{6}}]$

$= 50 + 25 + 12 + 6 + 3 + 1 = 97$

Exponent of $2$ in $(50)!$ is
$[\frac{50}{2}] + [\frac{50}{2^{2}}] + [\frac{50}{2^{3}}] + \dots + [\frac{50}{2^{5}}]$

$= 25 + 12 + 6 + 3 + 1 = 47$
Hence, exponent of $2$ in $^{100} C_{50}$ is $3.$
${∵^{100} C_{50} = \frac{2^{97}}{2^{47} \cdot 2^{47}} I = 2^{3} I}$

Exponent of $3$ in $(100)!$
$[\frac{100}{3}] + [\frac{100}{3^{2}}] + [\frac{100}{3^{3}}] + [\frac{100}{3^{4}}]$

$= 33 + 11 + 3 + 1 = 48$

Exponent of $3$ in $(50)!$ is
$[\frac{50}{3}] + [\frac{50}{3^{2}}] + [\frac{50}{3^{3}}] + [\frac{50}{3^{4}}]$

$= 16 + 5 + 1 = 22$
Hence, exponent of $3$ in $^{100} C_{50}$ is $4.$

Exponent of $5$ in $(100)!$ is
$[\frac{100}{5}] + [\frac{100}{5^{2}}] + [\frac{100}{5^{3}}]$

$= 20 + 4 + 0 = 24$

Exponent of $5$ in $(50)!$ is
$[\frac{50}{5}] + [\frac{50}{5^{2}}] + [\frac{50}{5^{3}}]$

$= 10 + 2 + 0 = 12$
Hence, exponent of $5$ in $^{100} C_{50}$ is $0.$

Exponent of $7$ in $(100)!$ is
$[\frac{100}{7}] + [\frac{100}{7^{2}}] + [\frac{100}{7^{3}}]$

$= 14 + 2 + 0 = 16$

Exponent of $7$ in $(50)!$ is
$[\frac{50}{7}] + [\frac{50}{7^{2}}] + [\frac{50}{7^{3}}]$

$= 7 + 1 + 0 = 8$
Hence, exponent of $7$ in $^{100} C_{50}$ is $0.$

Exponent of $2, 3, 5, 7$ in $^{100} C_{50}$ are $3, 4, 0, 0$ respectively.
$\Rightarrow α = 3, β = 4, γ = 0, δ = 0$

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If 100C50 can be prime factorised as 2α⋅3β⋅5γ⋅7δ… where α, β, γ, δ,… are non-negative integers, then the CORRECT relation(s) is (are)

If $^{100} C_{50}$ can be prime factorised as $2^{α} \cdot 3^{β} \cdot 5^{γ} \cdot 7^{δ} \dots$ where $α, β, γ, δ, \dots$ are non-negative integers, then the CORRECT relation(s) is (are)