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Permutations and Combinations question

2023 · 25 Jan · Shift 2 · Q37
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Permutations and Combinations question

2023 · 25 Jan · Shift 2 · Q37

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
∑k=0651−kC3\sum\limits_{k = 0}^6 {{}^{51 - k}{C_3}}k=0∑6​51−kC3​ is equal to :
  1. A
    51C4−45C4\mathrm{{}^{51}{C_4} - {}^{45}{C_4}}51C4​−45C4​
  2. B
    51C3−45C3\mathrm{{}^{51}{C_3} - {}^{45}{C_3}}51C3​−45C3​
  3. C
    52C3−45C3\mathrm{{}^{52}{C_3} - {}^{45}{C_3}}52C3​−45C3​
  4. D
    52C4−45C4\mathrm{{}^{52}{C_4} - {}^{45}{C_4}}52C4​−45C4​
View written solutionFree

Correct answer: D

  1. Rewrite the sum clearly

Given

∑k=0651−kC3\sum_{k=0}^{6} {^{51-k}C_3}k=0∑6​51−kC3​

When k=0,1,2,…,6k=0,1,2,\dots,6k=0,1,2,…,6, the upper index becomes:

51,50,49,48,47,46,4551,50,49,48,47,46,4551,50,49,48,47,46,45

So the sum is

51C3+50C3+49C3+48C3+47C3+46C3+45C3{^{51}C_3}+{^{50}C_3}+{^{49}C_3}+{^{48}C_3}+{^{47}C_3}+{^{46}C_3}+{^{45}C_3}51C3​+50C3​+49C3​+48C3​+47C3​+46C3​+45C3​
  1. Use the standard identity

We use:

∑r=mnrCt=n+1Ct+1−mCt+1\sum_{r=m}^{n} {^rC_t} = {^{n+1}C_{t+1}} - {^mC_{t+1}}r=m∑n​rCt​=n+1Ct+1​−mCt+1​

This follows from the hockey-stick identity:

∑r=tnrCt=n+1Ct+1\sum_{r=t}^{n} {^rC_t} = {^{n+1}C_{t+1}}r=t∑n​rCt​=n+1Ct+1​

Here, t=3t=3t=3, m=45m=45m=45, n=51n=51n=51. Therefore,

∑r=4551rC3=52C4−45C4\sum_{r=45}^{51} {^rC_3} = {^{52}C_4} - {^{45}C_4}r=45∑51​rC3​=52C4​−45C4​
  1. Match with the options

Thus,

∑k=0651−kC3=52C4−45C4\sum_{k=0}^{6} {^{51-k}C_3} = {^{52}C_4} - {^{45}C_4}k=0∑6​51−kC3​=52C4​−45C4​

This is Option D.

  1. Verification of other options
  • Option A: 51C4−45C4 {^{51}C_4} - {^{45}C_4}51C4​−45C4​ misses the required upper term 52C4^{52}C_452C4​.
  • Option B: wrong combination degree.
  • Option C: wrong combination degree.
  • Option D: correct.

Hence the correct answer is:

52C4−45C4\boxed{{^{52}C_4 - {}^{45}C_4}}52C4​−45C4​​
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