Given:
We have toys with bearing 4, 4 and 5
To Find: Number of 3-digit numbers he can make
The formula used: The number of permutations of n objects, where p1 objects are of one kind, p2 are of the second kind, ..., pk is of a kth kind and the rest, if any, are of a
different kind is = \(\frac{n!}{P_1!p_2! ......P_k!}\)
The child has to form a 3-digit number.
Here the child has two 4’s.
We have to use the above formula
Where,
n = 3
p1 = 2
\(\frac{3!}{2!}\) = 3 ways
The numbers are 544, 454 and 445.
He can make 3 3-digit numbers.
