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MHTCET 23 Apr 2025 Shift 2 Paper

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Question : 149 of 150

Marks: +1, -0

If

{ }^{15} C_{4}+{ }^{15} C_{5}+{ }^{16} C_{6}+{ }^{17} C_{7}+{ }^{18} C_{8}={ }^{19} C_{\mathrm{r}}

, then the value of

r

is equal to

9 or 10
7 or 12
8 or 10
8 or 11

Solution:

Concept:

Use Pascal's identity:

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

.

Explanation:

Combine the first two terms using the identity.

{ }^{15}C_{4} + { }^{15}C_{5} = { }^{16}C_{5}

.

So the sum becomes

{ }^{16}C_{5} + { }^{16}C_{6} + { }^{17}C_{7} + { }^{18}C_{8}

.

Now combine

{ }^{16}C_{5} + { }^{16}C_{6} = { }^{17}C_{6}

.

The expression reduces to

{ }^{17}C_{6} + { }^{17}C_{7} + { }^{18}C_{8}

.

Again,

{ }^{17}C_{6} + { }^{17}C_{7} = { }^{18}C_{7}

.

Now we have

{ }^{18}C_{7} + { }^{18}C_{8} = { }^{19}C_{8}

.

Hence

{ }^{19}C_{8} = { }^{19}C_{r}

.

Since

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

, either

r = 8

or

r = 19 - 8 = 11

.

Answer:

r = 8

or

11

Correct option: D.

8

or

11

.

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