The number of sequences of ten terms, whose terms are either 0 or 1 or 2 , that contain exactly five 1 s and exactly three 2 s , is equal to :

Solution

Maths sir · 4 steps

Five 1s and three 2s leave two 0s, so arrange ten objects in which the repeated digits look alike.

1. Count the zeros. There are 10 terms in all. Five are 1s and three are 2s, so the remaining 10 - 5 - 3 = 2 terms must be 0s.
2. Arrange with repeats. Every sequence is one arrangement of the ten digits 1, 1, 1, 1, 1, 2, 2, 2, 0, 0. Swapping two identical digits gives the same sequence, so divide 10! by the factorial of each repeat count.

   10!/(5! 3! 2!) = 3628800/(120 × 6 × 2) = 3628800/1440 = 2520

3. Check by positions. Pick the 5 places for the 1s in C(10, 5) = 252 ways. Then pick 3 of the 5 places left for the 2s in C(5, 3) = 10 ways. The 0s fill the rest, giving 252 × 10 = 2520.
4. So. There are 2520 such sequences, which is option (b).

The idea to keep. To arrange n items where some repeat, use n!/(p! q! r!). Always work out the leftover group first, here the 0s, because it is a repeat too.

Answer from the NTA final answer key. Solution by PrepWiser.