UPSC NDA I 2024 Mathematics Paper

Concept:We need to select 3 women and 3 men from the two groups, ensuring exactly 3 relatives come from the man's side and 3 from the wife's side.Explanation:Consider the possible composition of the 3 people taken from the man's side.Let the number of women from the man's side be $x$. Then the number of men from his side is $3-x$.The wife's side must supply the remaining $3-x$ women and $x$ men to satisfy the total of 3 women and 3 men.Case 1: $x=3$ (3 women from man's side, 0 men).Man's side: $\mathrm{C}(4,3) \times \mathrm{C}(3,0) = 4 \times 1 = 4$Wife's side: $\mathrm{C}(3,0) \times \mathrm{C}(4,3) = 1 \times 4 = 4$Ways: $4 \times 4 = 16$Case 2: $x=0$ (0 women from man's side, 3 men).Man's side: $\mathrm{C}(4,0) \times \mathrm{C}(3,3) = 1 \times 1 = 1$Wife's side: $\mathrm{C}(3,3) \times \mathrm{C}(4,0) = 1 \times 1 = 1$Ways: $1 \times 1 = 1$Case 3: $x=2$ (2 women, 1 man from man's side).Man's side: $\mathrm{C}(4,2) \times \mathrm{C}(3,1) = 6 \times 3 = 18$Wife's side: $\mathrm{C}(3,1) \times \mathrm{C}(4,2) = 3 \times 6 = 18$Ways: $18 \times 18 = 324$Case 4: $x=1$ (1 woman, 2 men from man's side).Man's side: $\mathrm{C}(4,1) \times \mathrm{C}(3,2) = 4 \times 3 = 12$Wife's side: $\mathrm{C}(3,2) \times \mathrm{C}(4,1) = 3 \times 4 = 12$Ways: $12 \times 12 = 144$Total ways: $16 + 1 + 324 + 144 = 485$Answer:485