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Q. Number of ways in which 6 different toys can be distributed equally among 3 children is also equal to

Permutations and Combinations

Solution:

Answer is $\left(\frac{6 !}{2 ! 2 ! 2 ! 3 !}\right) \times 3 !=\frac{6 !}{2 ! 2 ! 2 !}=90$
(A) $\frac{6 !}{2 ! 2 ! 2 !} \Rightarrow$ (A) is correct.
(B) ${ }^{10} C _2 \cdot 2 ! \Rightarrow$
(B) is correct.
image
(Using Division and Distribution)
(D) A and $S$ remains in $2^{\text {nd }}$ and $8^{\text {th }}$ position
Hence, number of ways $=\frac{6 !}{2 ! 2 ! 2 !} \Rightarrow$(D) is correct.