vinay1983 wrote:How many different 9 letter sequences can be formed such that they contain 2 Ps, 3Bs and 4Cs?
1. 1300
2. 1320
3. 1260
4. 1340
5. 1420
The number of ways to arrange 9 DISTINCT elements = 9!.
But the arrangement here includes IDENTICAL elements: 2 P's, 3 B's, and 4 C's.
When an arrangement includes IDENTICAL elements, we must divide by the number of ways EACH SET of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! (to account for the 2 P's), by 3! (to account for the 3 P's), and by 4! (to account for the 4 C's):
9!/(2!3!4!) = 1260.
The correct answer is
C.
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