Q155 / 270 · NDA 2024 (Phase 1) · Mathematics

A man has 7 relatives ( 4 women and 3 men). His wife also has 7 relatives ( 3 women and 4 men). In how many ways can they invite 3 women and 3 men so that 3 of them are man's relatives and 3 of them are his wife's relatives?

  • A
    340
  • B
    484
  • Correct answer
    C
    485
  • D
    469

Solution

Total number of ways $\begin{aligned} &={ }^4 \mathrm{C}_3 \times{ }^4 \mathrm{C}_3=4 \times 4=16 \\ &+{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_1 \\ &=18 \times 18=324 \\ &+{ }^3 \mathrm{C}_3 \times{ }^4 \mathrm{C}_0 \times{ }^4 \mathrm{C}_0 \times{ }^3 \mathrm{C}_3 \\ &=1 \times 1 \times 1 \times 1=1 \\ &+{ }^3 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2 \\ &=12 \times 12=144 \end{aligned}$ Hence, they can invite total no. of ways $=16+324+1+144=485 .$

Hence, the correct answer is option: 'C'.

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