CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

$$ ^{ 47 }C_{ 4 }$$ +$$ \sum _{ r=1 }^{ 5 }$$ .$$^{ 52-r }C_{ 3 } $$is equal to :


A
51C4
loader
B
52C4
loader
C
53C4
loader
D
None of these
loader

Solution

The correct option is B $$ ^{52}C_4 $$
$$ = ^{47}C_4+ ^{ 52-1}C_3 + ^{52-2}C_3 + ^{52-3}C_3 + ^{52-4}C_3 +^{52-5}C_3 $$
$$ =^{47}C_4 + ^{51}C_3 + ^{50}C_3 + ^{49}C_3 +^{4}C_3 + ^{47}C_3 $$
$$ =(^{47}C_4 + ^{47}C_3) + ^{4}C_3 +^{49}C_3 +^{50}C_3 +^{51}C_3 $$
$$ = (^{48}C_4 +^{48}C_3 + ^{49}C_3 +^{50}C_3 +^{51}C_3 $$
$$( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r) $$
$$ = (^{49}C_4 +^{49}C_3) + ^{50}C_3 + ^{51}C_3)( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r) $$
$$ =(^{50}C_4 + ^{50}C_3) + ^{51}C_3 )(\because ^nC_r +^nC_{r-1} = ^{n+1}C_r) $$
$$ =(^{51}C_4 + ^{51}C_3 \quad ( \because ^nC_r +^nC_{r-1} = ^{n+1}C_r )$$
$$ =^{52}C_4 \quad ( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r )$$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image