4 votes

Number of ways to form a sequence of 10 letters from four 'a's, four 'b's, four 'c's, and four 'd's if each letter must appear at least twice

The original poster wanted a solution with generating functions. So here is my take on that: I would use ordinary generating function instead of exponential one but still write: $ A= \frac{x^{4}}{24} +...
mond's user avatar
  • 41
3 votes
Accepted

polynomial solution of $a_n + ca_{n-1} + a_{n-2} = 0$

Hypothesis: $a_n=\sum_{i=0}^M b_i n^i$, $b_M \ne 0$. It's simpler to write the recurrence as $a_{n+1} + ca_{n} + a_{n-1} = 0$ which leads to $\sum_{i=0}^M b_i (n+1)^i + c\sum_{i=0}^M b_i n^i + \sum_{i=...
Alex K's user avatar
  • 1,529
4 votes
Accepted

Generate number with odd number of divisors

All the divisors of a natural number $n$ are split into pairs in the following way: $a$ and $b$ go in pair if $ab=n$. Since all the divisors are in pairs, there is an even number of them. When would ...
Aig's user avatar
  • 2,350
3 votes
Accepted

Number of ways to form a sequence of 10 letters from four 'a's, four 'b's, four 'c's, and four 'd's if each letter must appear at least twice

Can I do it directly using permutation with limited repetition? Sure. Every letter has to appear twice. That means $8$ out of the $10$ positions are already occupied and we have only $2$ positions ...
Haris's user avatar
  • 2,632
2 votes

Number of ways to form a sequence of 10 letters from four 'a's, four 'b's, four 'c's, and four 'd's if each letter must appear at least twice

You can do it with basic combinatorial methods: What you have are the letters aabbccdd?? with ?? representing wild card letters. There are $\frac{8!}{2!2!2!2!}$. There are 4 possibilities where the ...
yolo's user avatar
  • 454
2 votes

Balls out of a sack

There are $\binom{15}6$ ways to choose any $6$ balls from the $15=4+5+6$ balls that we have. How many ways to choose $6$ balls in such way that at least $1$ is red and at least $2$ are yellow? Let us ...
Aig's user avatar
  • 2,350
1 vote
Accepted

Extracting formula by pattern of child's game

First of all, let's split the problem into two parts. (1) What is the correspondence between position in the sequence and letter? (2) What is the correspondence between position in your rectangular ...
Gareth McCaughan's user avatar
1 vote

Number of ways to form a sequence of 10 letters from four 'a's, four 'b's, four 'c's, and four 'd's if each letter must appear at least twice

Since two different answers have been posted, here is a mechanical way of computing as the product of two multinomial coefficients, viz. $\mathtt{(Lay\;Down\;Pattern)\times(Permute\; pattern)}$ $ = \...
true blue anil's user avatar

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