3

The number of ways of selecting 4 letters out of the letters MANIMAL

A. 16
B. 17
C. 18
D. 19

I have made three different cases. Including 1 M, 2 M and none of the M.

So it is 6C3/2 + 5C3 + 4C3 which doesn't meet any of the option.

Freddy
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    Is it 4 letters or 4 pairwise distinct letters? – Stefan Mesken Jun 21 '15 at 05:48
  • @stefan there should be 4 distinct letters – Freddy Jun 21 '15 at 06:04
  • If it's $4$ distinct letters then the answer is $\binom54=5$. If it's $4$ non-distinct letters, I counted $22$ combinations. – barak manos Jun 21 '15 at 06:05
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    BTW, you need to change either the title of the question or the body of the question. The title says "$4$ letter words", which implies that the order of the letters matters (as is the case with words). The body of the question says "selecting $4$ letters", which implies that the order of the letters does not matter (as is the case with selections). – barak manos Jun 21 '15 at 06:11
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    Does this answer your question? How to find the number of $k$-permutations of $n$ objects with $x$ types, and $r_1, r_2, r_3, \cdots , r_x$ = the number of each type of object? –  Feb 17 '23 at 11:50
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    @user35395 Not a duplicate; linked question counts ordered selections, this question is about unordered selections. – Mike Earnest Feb 17 '23 at 18:21
  • @MikeEarnest What would be the UNordered equivalent of the linked question, in terms of $k, n, x, r_1, \cdots, r_x$? How would you write it? –  Feb 18 '23 at 19:46
  • @user35395 Essentially this: https://math.stackexchange.com/questions/553960/ If there are $k$ varieties of object, and such that there are $r_1,\dots,r_k$ units of each object available, then the number of ways to make an UNordered selection of $n$ objects is the number of solutions to the integer equation $x_1+\dots+x_k=n$ for which $0\le x_i\le r_i$ for each $i\in {1,\dots,k}$. The value of $x_i$ represents how many objects were taken of the $i^\text{th}$ type. Since this is UNordered, only the number of each object taken matters. – Mike Earnest Feb 18 '23 at 19:51
  • You mean, how did I know the MANIMAL question was unordered? I agree it was unclear based on the title, and the question statement. I inferred this based on the comments, and the multiple choice answers given. – Mike Earnest Feb 18 '23 at 19:56
  • @MikeEarnest thanks, yes. thanks. You wrote "$k$ varieties of object", but are you straying from the variables as defined in the linked question? The linked question had $n$ objects with $x$ types. Changing variables now can baffle other readers. By the way, do you want to close this as duplicate of https://math.stackexchange.com/q/553960? –  Feb 18 '23 at 19:58
  • I changed the notation because I wanted to use $x$ for the variable name instead of the subscript. $\tag*{}$I do not think this needs to be closed as a dupe, but I would not be mad if it was. The question with 553960 in its url is very general, requiring heavy duty methods like generating functions or principle of inclusion-exclusion, while this question can be solved by direct counting. Both perspectives are useful, so I think this question has some value not provided by Q553960. – Mike Earnest Feb 18 '23 at 20:05
  • @MikeEarnest How did you know that https://math.stackexchange.com/questions/553960 is UNordered, when it never mooted "order"? Because it mentioned Stars and Bars? $\tag*{}$ Similarly, how did you know that https://math.stackexchange.com/questions/2372/how-to-find-the-number-of-k-permutations-of-n-objects-with-x-types-and-r is ordered? Because it mentions "permutations"? –  Feb 19 '23 at 09:54
  • For the second question, yes, permutations means ordered selections. For the first question, I just have enough experience in combinatorics to know that counting solutions to equations like $x_1+\dots+x_k=n$ is useful for counting unordered selections. This will be my last reply to this comment chain. – Mike Earnest Feb 19 '23 at 17:32

4 Answers4

5

Hint $$(1+x+x^2)^2(1+x)^3=x^7+5 x^6+12 x^5+\color{red}{18} x^4+{18} x^3+12 x^2+5 x+1$$

user84413
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RE60K
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4

Add up the following, and you get $18$:

  • The number of combinations with $A=0$ and $M=1$ is $\binom33=1$
  • The number of combinations with $A=0$ and $M=2$ is $\binom32=3$
  • The number of combinations with $A=1$ and $M=0$ is $\binom33=1$
  • The number of combinations with $A=1$ and $M=1$ is $\binom32=3$
  • The number of combinations with $A=1$ and $M=2$ is $\binom31=3$
  • The number of combinations with $A=2$ and $M=0$ is $\binom32=3$
  • The number of combinations with $A=2$ and $M=1$ is $\binom31=3$
  • The number of combinations with $A=2$ and $M=2$ is $\binom30=1$
barak manos
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1

the answer should be for MNIMAMAL.,

$(1+x)^3(1+x+x^2)(1+x+x^2+x^3)$ $=$ $x^8+5x^7+12x^6+19x^5+22x^4+19x^3+12x^2+5x+1$

hence,the answer will be 22

TIWARI
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  • MNIMAMAL is the original word that OP has posted. I also calculated $22$ (see my comment to the question), but then OP edited the question twice and eventually changed it to MANIMAL. But all of this was about $2$ hours ago, so it's not really clear why you chose to answer it after OP had already fixed it. – barak manos Jun 21 '15 at 08:18
  • Nevertheless, this answer should not be down-voted, since it refers to the original question (and answers it correctly). Whoever down-voted it, obviously did not bother to read the comment above (or to understand why you chose to refer to the word 'MNIMAMAL' instead). – barak manos Jun 21 '15 at 09:15
0

We can consider the number of pairs of repeated letters:

1) If there are no repeated letters, we have $\dbinom{5}{4}=5$ selections.

2) If there is one pair of repeated letters, we have $\dbinom{2}{1}\dbinom{4}{2}=12$ selections.

3) If there are 2 pairs of repeated letters, we have only 1 selection.

Therefore there are 18 possible selections altogether.

user84413
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