The value of { }^{47} C₄+_{{j}=1}^{5}{ }^{(52-{j})} C₃ is

Correct Answer is : 52C4{ }^{52} C₄52C4

Correct Answer is : 52C4{ }^{52} C₄52C4

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MHTCET 5 May 2025 Shift 2 Paper

Question : 110 of 150

Marks: +1, -0

The value of

{ }^{47} C_{4}+\sum_{\mathrm{j}=1}^{5}{ }^{(52-\mathrm{j})} C_{3}

Solution:

Concept:

Use Pascal’s identity:

{ }^{n}C_{r} + { }^{n}C_{r-1} = { }^{n+1}C_{r}

Explanation:

Write the given expression in expanded form:

{ }^{47}C_{4} + { }^{51}C_{3} + { }^{50}C_{3} + { }^{49}C_{3} + { }^{48}C_{3} + { }^{47}C_{3}

Group

{ }^{47}C_{4}

and

{ }^{47}C_{3}

{ }^{47}C_{4} + { }^{47}C_{3} = { }^{48}C_{4}

Now combine

{ }^{48}C_{4} + { }^{48}C_{3} = { }^{49}C_{4}

Next,

{ }^{49}C_{4} + { }^{49}C_{3} = { }^{50}C_{4}

Then,

{ }^{50}C_{4} + { }^{50}C_{3} = { }^{51}C_{4}

Finally,

{ }^{51}C_{4} + { }^{51}C_{3} = { }^{52}C_{4}

Thus, the whole sum simplifies to

{ }^{52}C_{4}

Answer:

{ }^{52}C_{4}

, which is option A.

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