Case 1: Both Ts appear. So we know that two of the five letters are Ts, so we need to choose the remaining 3 letters from the 10 available. [Math Processing Error](103)=120. Then we need to arrange the 5 letters. There are [Math Processing Error]5!=120 ways to arrange 5 things. However, since we know we have two identical Ts we’ll find that 5! gives each word “twice” (ie, it treats the Ts as T1 and T2 and so you get both arrangements of Ts). So we need to divide by 2: [Math Processing Error]5!2=1202=60 ways to sort the 5 letters with two Ts. So there are total [Math Processing Error]120∗60=7200 five-letter permutations in which both Ts are featured.
Case 2: At most one T appears. Since it doesn’t matter which T (they’re identical), we can essentially just set one aside, and that guarantees that we don’t get them both. So we’re left with 11 letters and need to pick out and arrange 5. There are actually 3 ways that come to mind to do this.
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Answer:
Case 1: Both Ts appear. So we know that two of the five letters are Ts, so we need to choose the remaining 3 letters from the 10 available. [Math Processing Error](103)=120. Then we need to arrange the 5 letters. There are [Math Processing Error]5!=120 ways to arrange 5 things. However, since we know we have two identical Ts we’ll find that 5! gives each word “twice” (ie, it treats the Ts as T1 and T2 and so you get both arrangements of Ts). So we need to divide by 2: [Math Processing Error]5!2=1202=60 ways to sort the 5 letters with two Ts. So there are total [Math Processing Error]120∗60=7200 five-letter permutations in which both Ts are featured.
Case 2: At most one T appears. Since it doesn’t matter which T (they’re identical), we can essentially just set one aside, and that guarantees that we don’t get them both. So we’re left with 11 letters and need to pick out and arrange 5. There are actually 3 ways that come to mind to do this.
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