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4 letters must be chosen from the word REMEMBRANCE

How many different selections can be done if there is no M, no R and at least 2 Es?

I've seen this question somewhere but i don't remember the question well

i know that the answer is either:

(3c2 x 4c2 ) + (3c3 x 4c1) = 22

or

(4c2 + 4c1) = 10

i am a bit confused can someone please tell me what is the right answer?

user105802
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  • Please refer to the mathJax tutorial found at the right-hand side of the page. – Don Larynx Nov 06 '13 at 06:44
  • The Question is a little ambiguous (what's the right answer?) because it's unclear whether "different selections" (in the body of the Question) refers to combinations (used in title) or permutations (used as tag). Keep in mind combinations do not treat the order (of letters) as significant, but permutations do treat the order as significant. Since you say you "don't remember the question well", it's possible you really do know which was originally intended, or that originally both values were required. – hardmath Nov 06 '13 at 16:09
  • i was not allowed to tag combination that is why i tagged permutations – user105802 Nov 06 '13 at 18:14

2 Answers2

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Well, since we are not supposed to select any M or R, then our selection pool is narrowed to $$\text{EEBANCE},$$ from which we are required to select two Es, and any two other letters. Our answer, then, is the number of ways we can select two letters from $$\text{BANCE}.$$ How many ways can this be done?

Cameron Buie
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Since the instructions are not to use any R or M, the first thing I'd do is omit them from the stock of available letters:

REMEMBRANCE --> EEBANCE

Now the problem can be done (tediously) by writing out all possibilities. To simplify matters, let's apply the final constraint of using at least two E's. Since there are only three E's available, that means two cases: we use exactly two E's or we use exactly three E's.

The case where three E's are used (to create a "word" of four letters) is especially easy to number. Where does the non-E character go in the word (permutations with repetitions), or if you don't consider order important (combinations with repetitions), simply how may ways are there to choose a letter that is not E?

The case where two E's are used should be tractable after you've done the warm-up exercise with three E's.

hardmath
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