Question

Answer the following by appropriately matching the lists based on the information given in the paragraph.

Let L (n, r) denotes the number of ways to select 'r' objects from 'n' distinct objects placed in a row, such that no two of the selected objects were consecutive, and let R (n, r) denotes the number of ways to select 'r' objects from 'n' distinct objects placed on a circle, such that no two of the selected objects were consecutive.

List-I (I) R(12, 4) + L(8, 2) = (II) R(14, 5) + L(10, 3) = (III) L(12, 4) = (IV) L (13, 5) =

List-II (P) $^{10}C_5$ (Q) $^8C_4 + ^8C_3$ (R) $^9C_5$ (S) $^8C_4$ (T) $^{11}C_5 - ^{10}C_4$

Which of the following is / are correct :

• A. I→Q, R
• B. II→Q, R
• C. III→S
• D. IV→ P,Q,R

Answer 1

Answer: (A)

Option (A) is correct:

• (I) → Q, R.

Answer 2

We use the formulas:

$L(n,r)=\binom{n-r+1}{r}$

and for a circle,

$R(n,r)=\frac{n}{n-r}\binom{n-r}{r}$.

For (I):

$L(8,2)=\binom{8-2+1}{2}=\binom{7}{2}=21$.

$R(12,4)=\frac{12}{12-4}\binom{12-4}{4}=\frac{12}{8}\binom{8}{4}=\frac{12}{8}\times70=105$.

Thus,

$R(12,4)+L(8,2)=105+21=126$.

Also,

Option Q: $\binom{8}{4}+\binom{8}{3}=70+56=126$,

Option R: $\binom{9}{5}=126$.

So (I) matches both Q and R.

For (II):

$L(10,3)=\binom{10-3+1}{3}=\binom{8}{3}=56$.

$R(14,5)=\frac{14}{14-5}\binom{14-5}{5}=\frac{14}{9}\binom{9}{5}=\frac{14}{9}\times126=196$.

Thus,

$R(14,5)+L(10,3)=196+56=252$,

which does not match Q (126) or R (126).

For (III):

$L(12,4)=\binom{12-4+1}{4}=\binom{9}{4}=126$,

but Option S is $\binom{8}{4}=70$. So no match.

For (IV):

$L(13,5)=\binom{13-5+1}{5}=\binom{9}{5}=126$,

and among List-II, only Q and R equal 126 while P is $\binom{10}{5}=252$; hence (IV) does not match P, Q, and R all.

Conclusion: Only Option (A) is correct:

• (I) → Q, R.

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Core Explanation (Minimal):
Compute:

• $L(8,2)=21$ and $R(12,4)=105$, so (I)=126, matching both Q ($70+56$) and R ($126$).
• Others do not match the corresponding list values.