The number of ways of dividing 15 persons into 3 groups containing 3,5 and 7 persons so that two particular persons are not included into the 5 persons group is

• Correct answer
  A

  $\frac{117(11!)}{3!(7!)}$
• B

  ${ }^{15} \mathrm{C}_5{ }^{10} \mathrm{C}_3$
• C

  $90 \times \frac{13!}{7!}$
• D

  ${ }^{15} \mathrm{C}_5{ }^8 \mathrm{C}_3$

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