Number of ways in which 9 different toys be distributed among 4 children belonging to different age groups in such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more is

Question

Filo · Accepted Answer

According to given question elder children will get 2 toys each and younger will get 3 toys.

No. of ways younger one will get 3 toys are

_{3}^{9} C = \frac{9 !}{3 ! \times 6 !}

No. of ways by which elder children will get 2 toys each

_{2}^{6} C \times_{2}^{4} C \times_{2}^{2} C = \frac{6 !}{2 ! \times 4 !} \times \frac{4 !}{2 ! \times 2 !} \times 1

Total ways taking both conditions together

= \frac{9 !}{3 ! \times 6 !} \times \frac{6 !}{2 ! \times 4 !} \times \frac{4 !}{2 ! \times 2 !} \times 1

= \frac{9 !}{3 ! ( 2 !) ^{3}}

Question Text	Number of ways in which $9$ different toys be distributed among $4$ children belonging to different age groups in such a way that distribution among the $3$ elder children is even and the youngest one is to receive one toy more is
Answer Type	Text solution:1
Upvotes	150

Number of ways in which $9$ different toys be distributed among $4$ children belonging to different age groups in such a way that distribution among the $3$ elder children is even and the youngest one is to receive one toy more is

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