Write the Value of 6 ∑ R = 1 56 − R C 3 + 50 C 4 - Mathematics

Write the value of$\sum^6_{r = 1} \ ^{56 - r}{}{C}_3 + \ ^ {50}{}{C}_4$

Solution

We know:
nC$-$1 + nCr = n+1Cr

$\text{Now, we have}:$
$\sum^6_{r = 1} {}^{56 - r} C_3 + {}^{50} C_4$
$=^{55} C_3 +^{54} C_3 +^{53} C_3 +^{52} C_3 +^{51} C_3 +^{50} C_3 +^{50} C_4$
$=^{55} C_3 +^{54} C_3 +^{53} C_3 +^{52} C_3 +^{51} C_3 +^{51} C_4$
$=^{55} C_3 +^{54} C_3 +^{53} C_3 +^{52} C_3 +^{52} C_4$
$=^{55} C_3 +^{54} C_3 +^{53} C_3 +^{53} C_4$
$=^{55} C_3 +^{54} C_3 +^{54} C_4$
$=^{55} C_3 +^{55} C_4$
$=^{56} C_4$
Concept: Factorial N (N!) Permutations and Combinations
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Q 5 | Page 24