TS EAMCET · Maths · Permutation Combination

Three and four digit numbers are formed from the digits $1, 3, 5, 6, 8 .$ If $e_{1}$ is number of three digit even numbers with no digit repeated and $e_{2}$ is number of four digit even numbers with no digit repeated. Also $O_{1}$ represent the number of three digit odd numbers in which no digit is repeated and $O_{2}$ represent the number of four digit odd numbers in which no digit is repeated. Then

A $e_{1} = O_{1}, e_{2} = O_{2}$
B $e_{1} + e_{2} + O_{1} + O_{2} =^{5} P_{3} + 5^{3}$
C $\frac{e_{1} + e_{2}}{2} = \frac{O_{1} + O_{2}}{3} = 6^{2}$
D $\frac{e_{1} + e_{2}}{O_{1} + O_{2}} = \frac{3}{2}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{e_{1} + e_{2}}{2} = \frac{O_{1} + O_{2}}{3} = 6^{2}$

Step-by-step Solution

Detailed explanation

∙e1 Number of ways=e1=3×4×2 ⇒e1=24 ∙e2 Number of ways =e2=2×3×4×2 ⇒e2=48 ∙O1 Number of ways=O1=3×4×3 ⇒O1=36 ∙O2 Number of ways =O2=2×3×4×3 ⇒O2=72 Now, e1+e22=36 =O1+O23=62

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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