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Worked solution

Distinguish separated Es from not all Es together

If the letters of SEPTEMBER are arranged at random, find the probability that (i) the 3 Es are together, (ii) the 3 Es are separated, (iii) not all the 3 Es are together.

CJC · Lecture · Example 6

Count equally likely distinct arrangements

There are 9 letters: S,P,T,M,B, Ronce each and 3 identical Es. The total number of distinct arrangements is 9!/3!=604809!/3!=60480. Uniformly shuffling the physical letter copies gives each distinct spelling the same multiplicity 3!, so the count ratio is valid.

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(i) Make one EEE block

The 7 outer objects are S,P,T,M,B,R,[EEE], arranged in 7!=5040 ways. Thus P(all Es together)=5040/60480=1/12P(\text{all Es together})=5040/60480=1/12. There is no extra internal order for identical Es.

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(ii) Put Es in three different gaps

Arrange the 6 other letters in 6! ways. They create 7 gaps including both ends. Choose 3 different gaps for the identical Es in (73)\binom73 ways. Hence 6!(73)=252006!\binom73=25200 separated arrangements and P(all Es separated)=25200/60480=5/12P(\text{all Es separated})=25200/60480=5/12.

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(iii) Complement only the all-together event

P(not all Es together)=1−P(all Es together)=1−1/12=11/12P(\text{not all Es together})=1-P(\text{all Es together})=1-1/12=11/12. This includes arrangements with exactly two adjacent Es. Follow the Teacher warning: not all together does not imply all separated, and 11/12 is not 5/12.

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Six source boxes for the six non-E letters, illustrated as S,P,T,M,B,R, and seven arrows into the gaps including both ends. Choose three distinct gaps for the three identical E copies. The order shown is illustrative; all six factorial orders are counted. No extra three factorial factor for the identical Es.
Six source boxes for the six non-E letters, illustrated as S,P,T,M,B,R, and seven arrows into the gaps including both ends. Choose three distinct gaps for the three identical E copies. The order shown is illustrative; all six factorial orders are counted. No extra three factorial factor for the identical Es.
Six source boxes for the six non-E letters, illustrated as S,P,T,M,B,R, and seven arrows into the gaps including both ends. Choose three distinct gaps for the three identical E copies. The order shown is illustrative; all six factorial orders are counted. No extra three factorial factor for the identical Es.

Common mistakes

The complement of all three together is not all three separated. Include the mixed case where two Es are together and the third is elsewhere. Count identical letters consistently in numerator and denominator.

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