A new December and a new bunch of puzzles from mscroggs.co.uk.
We’re looking for a 2 digit number, ab (and the product ab * ba). It can’t be a 1 digit number, a, because a * a wouldn’t have 3 digits. It can’t be a 3 digit number, abc, because abc * cba would have at least 5 digits.
Also, ab is a square. It could be 16, 25, 36, 49, 64, 81. Going through these manually shows 16 * 61 = 976 is the only 3 digit number.
This week the question is: Can You Survive Squid Game (Season 2)?
In Season 2 of Squid Game, contestants play a game of “Mingle” (spoiler alert!). For each round of this game, the contestants all wait in a common area until a number is called. They must then split up into groups of that number. Any contestants not in a group of the correct number after 30 seconds … well, let’s just say bad things happen to them. For now, we’ll refer to contestants who make it through a round in a group of the correct size as having “survived” the round.
Suppose there are initially N contestants.
In the first round, contestants must form groups of 1. Okay, that’s not too hard; everyone survives. In the second round, contestants must form groups of 2. This continues (with groups of k in round k) until the 38th round, when contestants must form groups of 38.
What is the smallest number N such that there is at least one person who survives all 38 rounds?
And for extra credit:
There are N contestants about to play Mingle, where N is less than 500. This time, in each round, the number for group size can be anywhere from 1 to 10, with each number equally likely to be called. Also, the numbers called in the rounds are all independent of each other. It appears this version of Mingle will continue until there are no more survivors.
Before the game starts, one of the contestants, Young-il, does some quick computation. He says to his companion, Gi-hun: “Oh no, it’s too bad we only have N people.” (Of course, Young-il doesn’t actually say “N,” but rather the number that N represents.)
Young-il continues: “If we had started with N+1 people instead, I’d expect the game to last about 10 more rounds, on average.”
What is the value of N?
Highlight to reveal (possibly incorrect) solution:
Program. A little messy.
If we begin with N persons, before round 1, of course, after round 1 there will be f(1) = N persons. After round 2, we need to take N, divide by 2 to get the number of groups, round down to nearest integer if necessary, and then multiply by 2. f(2) = ⌊ f(1)/2 ⌋ * 2. And it goes on like this: f(3) = ⌊ f(2)/3 ⌋ * 3. In general, after round M we have f(M) = ⌊ f(M-1)/M ⌋ * M.
I wrote a program to calculate f(38), given an N. Some trial and error showed, that given N = 500, f(38) > 0. I then rewrote the program to try all possible N from 1, and stop when f(38) was positive the first time. Solution: we were looking for N = 454.
Just for fun I wrote some more code. Going the opposite direction, we might say f(37) = ⌈ f(38)/37 ⌉ * 37. Basing a small algorithm on this, I got the same solution, but I wasn’t sure that wasn’t luck.
Part 2:
For this I wrote another program. This one really calls out for some Monte Carlo. Try all N from 1 to 500, simulate X executions, calculate the average length of an execution (how many rounds), run the program, export to csv, open in Libre Office Calc, make an extra column with the difference between lengths for this particular N and the one above, some conditional formatting to look for a difference close to 10, and bam! Just to make sure, I looked a little closer at this neighborhood, and it just became clearer, that I was looking in the right place (see PDF above). The solution is N = 359.
I have a suspicion, this is because 360 is such a nice number. Coming from 361, 362, 363 etc., there’s a good chance of landing here. And then only moving on to a lower number, if the random number is 7.
This December I participated in Advent of Code.
Part 1 is rather easy. Let the robots walk.
Part 2 proved much more challenging. How does one detect a picture? I developed a strategy of trying to sense, whether the robots were “clumping”, and then printing the map, to get a visual check. This worked, and I found the picture. Then a little while later I discovered, that the apparently random output from part 1 could be used to detect the picture, but I didn’t implement that fully. Still, I really liked that elegant solution. This version of the program was actually built partially after I know the solution.
I’m getting a bit more systematic. Having all the values, that change between test and actual data at the top.
Part 1:
<?php
$input = file_get_contents('./d14input2.txt', true);
$mapwid = 101;
$maphei = 103;
// test: 11 x 7
// actual: 101 x 103
$robots = array();
foreach(preg_split("/((\r?\n)|(\r\n?))/", $input) as $line) {
if(strlen($line)>2) {
preg_match_all('/-?\d+/', $line, $robot);
$robots[] = $robot[0];
}
}
function print_map() {
global $robots, $mapwid, $maphei;
$map = array();
foreach($robots as $robot) {
$x = $robot[0];
$y = $robot[1];
if(isset($map[$x][$y])) {
$map[$x][$y]++;
} else {
$map[$x][$y] = 1;
}
}
for($j=0;$j<$maphei;$j++) {
for($i=0;$i<$mapwid;$i++) {
if(isset($map[$i][$j])) {
echo $map[$i][$j];
} else {
echo ".";
}
}
echo "\n";
}
echo "\n";
}
for($i=0;$i<100;$i++) {
// print("$i seconds elapsed\n");
for($j=0;$j<sizeof($robots);$j++) {
$robots[$j][0] =
($robots[$j][0] + $robots[$j][2] + $mapwid) % $mapwid;
$robots[$j][1] =
($robots[$j][1] + $robots[$j][3] + $maphei) % $maphei;
}
}
$robx0y0 = 0;
$robx0y1 = 0;
$robx1y0 = 0;
$robx1y1 = 0;
for($j=0;$j<sizeof($robots);$j++) {
if($robots[$j][0] < ($mapwid-1)/2) {
if($robots[$j][1] < ($maphei-1)/2) {
$robx0y0++;
} else {
if($robots[$j][1] > ($maphei-1)/2) {
$robx0y1++;
}
}
} else {
if($robots[$j][0] > ($mapwid-1)/2) {
if($robots[$j][1] < ($maphei-1)/2) {
$robx1y0++;
} else {
if($robots[$j][1] > ($maphei-1)/2) {
$robx1y1++;
}
}
}
}
}
print("$robx0y0 $robx0y1 $robx1y0 $robx1y1\n");
$prod = $robx0y0 * $robx0y1 * $robx1y0 * $robx1y1;
print("$prod\n");
?>Part 2:
<?php
$input = file_get_contents('./d14input2.txt', true);
$mapwid = 101;
$maphei = 103;
// test: 11 x 7
// actual: 101 x 103
$robots = array();
foreach(preg_split("/((\r?\n)|(\r\n?))/", $input) as $line) {
if(strlen($line)>2) {
preg_match_all('/-?\d+/', $line, $robot);
$robots[] = $robot[0];
}
}
//print_r($robots);
function print_map() {
global $robots, $mapwid, $maphei;
$map = array();
foreach($robots as $robot) {
$x = $robot[0];
$y = $robot[1];
if(isset($map[$x][$y])) {
$map[$x][$y]++;
} else {
$map[$x][$y] = 1;
}
}
for($j=0;$j<$maphei;$j++) {
for($i=0;$i<$mapwid;$i++) {
if(isset($map[$i][$j])) {
echo $map[$i][$j];
} else {
echo ".";
}
}
echo "\n";
}
echo "\n";
}
function xmas_tst() {
global $robots, $mapwid, $maphei;
$robx0y0 = 0;
$robx0y1 = 0;
$robx1y0 = 0;
$robx1y1 = 0;
for($j=0;$j<sizeof($robots);$j++) {
if($robots[$j][0] < ($mapwid-1)/2) {
if($robots[$j][1] < ($maphei-1)/2) {
$robx0y0++;
} else {
if($robots[$j][1] > ($maphei-1)/2) {
$robx0y1++;
}
}
} else {
if($robots[$j][0] > ($mapwid-1)/2) {
if($robots[$j][1] < ($maphei-1)/2) {
$robx1y0++;
} else {
if($robots[$j][1] > ($maphei-1)/2) {
$robx1y1++;
}
}
}
}
}
if(max($robx0y0,$robx0y1,$robx1y0,$robx1y1) > 150) {
return 1;
} else {
return 0;
}
}
$handle = fopen ("php://stdin","r");
for($i=0;$i<8258;$i++) {
for($j=0;$j<sizeof($robots);$j++) {
$robots[$j][0] =
($robots[$j][0] + $robots[$j][2] + $mapwid) % $mapwid;
$robots[$j][1] =
($robots[$j][1] + $robots[$j][3] + $maphei) % $maphei;
}
if(xmas_tst()) {
print_map();
print($i+1 . " seconds elapsed\n");
$line = fgets($handle);
}
}
fclose($handle);
?>A new December and a new bunch of puzzles from mscroggs.co.uk.
Here we are helped by a very nice pattern for 15n.
Very quickly for odd n, the last 3 digits are 375, and for even n, the last 3 digits are 625. The n in the question is even, so the answer is 625.
This December I participated in Advent of Code.
Ah yes. It took me a little while to wrap my head around “2 equations with 2 variables”. Doh. And then it took me a little while to actually implement a standard solution. (Question.) Double doh.
For part 2 I had to change 2 lines, so below is only part 2.
Part 2:
<?php
$input = file_get_contents('./d13input2.txt', true);
$rules = array();
$i = 0;
foreach(preg_split("/((\r?\n)|(\r\n?))/", $input) as $line) {
if(preg_match("/Button A/",$line,$matches)) {
if(preg_match_all("/(\d+)/",$line,$matches)) {
$rules[$i]["ax"] = $matches[0][0];
$rules[$i]["ay"] = $matches[0][1];
}
}
if(preg_match("/Button B/",$line,$matches)) {
if(preg_match_all("/(\d+)/",$line,$matches)) {
$rules[$i]["bx"] = $matches[0][0];
$rules[$i]["by"] = $matches[0][1];
}
}
if(preg_match("/Prize/",$line,$matches)) {
if(preg_match_all("/(\d+)/",$line,$matches)) {
// part 1 doesn't have the +10000000000000
$rules[$i]["px"] = $matches[0][0] + 10000000000000;
$rules[$i]["py"] = $matches[0][1] + 10000000000000;
}
}
if(strlen($line) < 2) {
$i++;
}
}
$money = 0;
$hits = 0;
foreach($rules as $rule) {
// Coefficients of the equations
$a1 = $rule["ax"]; $b1 = $rule["bx"]; $c1 = $rule["px"];
$a2 = $rule["ay"]; $b2 = $rule["by"]; $c2 = $rule["py"];
// Calculate the determinant
$determinant = $a1 * $b2 - $a2 * $b1;
if ($determinant == 0) {
// The system has no unique solution
// (either no solution or infinitely many solutions)
} else {
// Calculate x and y using Cramer's Rule
$x = ($c1 * $b2 - $c2 * $b1) / $determinant;
$y = ($a1 * $c2 - $a2 * $c1) / $determinant;
}
$soln = $y;
$solm = $x;
if(abs((int)$soln - $soln) < 0.01
&& abs((int)$solm - $solm) < 0.01) {
if($solm > 0 && $soln > 0) {
$money += $solm * 3 + $soln;
$hits++;
}
}
}
print("$hits $money\n");
?>A new December and a new bunch of puzzles from mscroggs.co.uk.
Let the square be:
abc def ghi
Rules:
From these we can deduce these further rules:
Assume i = 0. Then by (1) f = 0. Then by (4) h = 0. Then by (1) e = 0. But by (3), then b = 19, not a digit. So, i != 0.
i = 5. Then by (1) f = 0. Then by (4), c05 = 3 gh5, this might be 405 = 3 * 135 or 705 = 3 * 235. In both cases h = 3. g might be 1 or 2. c might be 4 or 7. Then by (1) e = 7. Then by (3) b = 9.
So far we have:
a9c d70 g35
Assume g = 1. Then by (1) d = 2. Then by (2) a = 12, not a digit. So, g != 1.
g = 2. Then by (1) d = 4. Then by (2) a = 9. Then by (4) c = 7.
The sought after number, abc, is 997.
This December I participated in Advent of Code.
This one had me thinking.
Part 1:
<?php
$input = file_get_contents('./d12input1.txt', true);
foreach(preg_split("/((\r?\n)|(\r\n?))/", $input) as $line) {
if(strlen($line)>2) {
$map[] = str_split($line);
}
}
$mapx = sizeof($map);
$mapy = sizeof($map[0]);
// name possible areas
$names = array();
for($i=0;$i<sizeof($map);$i++) {
for($j=0;$j<sizeof($map[$i]);$j++) {
$names[$map[$i][$j]] = 1;
}
}
// convert names
$names2 = array();
foreach($names as $key => $val) {
$names2[] = $key;
}
$names = $names2;
// thin areas out
$areas = array();
for($k=0;$k<sizeof($names);$k++) {
$lett = $names[$k];
$tmp = $map;
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if($tmp[$i][$j] != $lett) {
unset($tmp[$i][$j]);
}
}
}
$areas[$k] = $tmp;
}
function subtst($i, $j) {
global $tmpb;
//print_r($tmpb);
if(isset($tmpb[$i-1][$j])) {
if($tmpb[$i-1][$j] != "*") {
$tmpb[$i-1][$j] = "*";
subtst($i-1, $j);
}
}
if(isset($tmpb[$i+1][$j])) {
if($tmpb[$i+1][$j] != "*") {
$tmpb[$i+1][$j] = "*";
subtst($i+1, $j);
}
}
if(isset($tmpb[$i][$j-1])) {
if($tmpb[$i][$j-1] != "*") {
$tmpb[$i][$j-1] = "*";
subtst($i, $j-1);
}
}
if(isset($tmpb[$i][$j+1])) {
if($tmpb[$i][$j+1] != "*") {
$tmpb[$i][$j+1] = "*";
subtst($i, $j+1);
}
}
}
function atst() {
global $tmp, $map, $tmpb, $key, $mapx, $mapy, $areas;
// grab random beginning point
$lett = "";
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if(isset($tmp[$i][$j])) {
$lett = $tmp[$i][$j];
$x = $i;
$y = $j;
break 2;
}
}
}
// can I reach all points from here?
$tmpb = $tmp;
$tmpb[$x][$y] = "*";
subtst($x, $y);
if(sizeof(array_count_values(array_merge(...$tmpb))) > 1) {
$new1 = $tmpb;
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if(isset($new1[$i][$j])) {
if($new1[$i][$j] == $lett) {
unset($new1[$i][$j]);
} else {
$new1[$i][$j] = $lett;
}
}
}
}
$new2 = $tmpb;
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if(isset($new2[$i][$j])) {
if($new2[$i][$j] != $lett) {
unset($new2[$i][$j]);
}
}
}
}
$areas[$key] = $new1;
$areas[] = $new2;
}
}
// if areas aren't actually connected, split up
$areaz = 0;
while($areaz < sizeof($areas)) {
$areaz = sizeof($areas);
foreach($areas as $key => $area) {
$tmp = $area;
atst();
}
}
$sum = 0;
foreach($areas as $key => $area) {
$sz1 = array_count_values(array_merge(...$area));
$sz2 = array_pop($sz1);
$per = 0;
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if(isset($area[$i][$j])) {
if(!isset($area[$i-1][$j])) {
$per++;
}
if(!isset($area[$i+1][$j])) {
$per++;
}
if(!isset($area[$i][$j-1])) {
$per++;
}
if(!isset($area[$i][$j+1])) {
$per++;
}
}
}
}
$sum += $sz2 * $per;
}
print("$sum\n");
?>Part 2:
<?php
$input = file_get_contents('./d12input2.txt', true);
foreach(preg_split("/((\r?\n)|(\r\n?))/", $input) as $line) {
if(strlen($line)>2) {
$map[] = str_split($line);
}
}
$mapx = sizeof($map);
$mapy = sizeof($map[0]);
// name possible areas
$names = array();
for($i=0;$i<sizeof($map);$i++) {
for($j=0;$j<sizeof($map[$i]);$j++) {
$names[$map[$i][$j]] = 1;
}
}
// convert names
$names2 = array();
foreach($names as $key => $val) {
$names2[] = $key;
}
$names = $names2;
// thin areas out
$areas = array();
for($k=0;$k<sizeof($names);$k++) {
$lett = $names[$k];
$tmp = $map;
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if($tmp[$i][$j] != $lett) {
unset($tmp[$i][$j]);
}
}
}
$areas[$k] = $tmp;
}
//print_r($areas);
function subtst($i, $j) {
global $tmpb;
if(isset($tmpb[$i-1][$j])) {
if($tmpb[$i-1][$j] != "*") {
$tmpb[$i-1][$j] = "*";
subtst($i-1, $j);
}
}
if(isset($tmpb[$i+1][$j])) {
if($tmpb[$i+1][$j] != "*") {
$tmpb[$i+1][$j] = "*";
subtst($i+1, $j);
}
}
if(isset($tmpb[$i][$j-1])) {
if($tmpb[$i][$j-1] != "*") {
$tmpb[$i][$j-1] = "*";
subtst($i, $j-1);
}
}
if(isset($tmpb[$i][$j+1])) {
if($tmpb[$i][$j+1] != "*") {
$tmpb[$i][$j+1] = "*";
subtst($i, $j+1);
}
}
}
function atst() {
global $tmp, $map, $tmpb, $key, $mapx, $mapy, $areas;
// grab random beginning point
$lett = "";
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if(isset($tmp[$i][$j])) {
$lett = $tmp[$i][$j];
$x = $i;
$y = $j;
break 2;
}
}
}
// can I reach all points from here?
$tmpb = $tmp;
$tmpb[$x][$y] = "*";
subtst($x, $y);
if(sizeof(array_count_values(array_merge(...$tmpb))) > 1) {
//print("alert\n");
$new1 = $tmpb;
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if(isset($new1[$i][$j])) {
if($new1[$i][$j] == $lett) {
unset($new1[$i][$j]);
} else {
$new1[$i][$j] = $lett;
}
}
}
}
$new2 = $tmpb;
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if(isset($new2[$i][$j])) {
if($new2[$i][$j] != $lett) {
unset($new2[$i][$j]);
}
}
}
}
$areas[$key] = $new1;
$areas[] = $new2;
}
}
// if areas aren't actually connected, split up
$areaz = 0;
while($areaz < sizeof($areas)) {
$areaz = sizeof($areas);
foreach($areas as $key => $area) {
$tmp = $area;
atst();
}
}
$sum = 0;
foreach($areas as $key => $area) {
$sz1 = array_count_values(array_merge(...$area));
$sz2 = array_pop($sz1);
$per = 0;
for($i=0;$i<$mapx;$i++) {
for($j=0;$j<$mapy;$j++) {
if(isset($area[$i][$j])) {
// above
if(!isset($area[$i-1][$j])) {
// already counted?
if(isset($area[$i][$j-1]) && !isset($area[$i-1][$j-1])) {
} else {
$per++;
}
}
// below
if(!isset($area[$i+1][$j])) {
// already counted?
if(isset($area[$i][$j-1]) && !isset($area[$i+1][$j-1])) {
} else {
$per++;
}
}
// to the left
if(!isset($area[$i][$j-1])) {
// already counted?
if(isset($area[$i-1][$j]) && !isset($area[$i-1][$j-1])) {
} else {
$per++;
}
}
// to the right
if(!isset($area[$i][$j+1])) {
// already counted?
if(isset($area[$i-1][$j]) && !isset($area[$i-1][$j+1])) {
} else {
$per++;
}
}
}
}
}
$sum += $sz2 * $per;
}
print("$sum\n");
?>A new December and a new bunch of puzzles from mscroggs.co.uk.
Holly picks the 3-digit number abc. Then she removes either a, b or c to get a new 2-digit number. Adding these 2 numbers together gets 309. 309 is odd, therefore removing a digit from abc can’t preserve c as the last digit, as that would make the sum even. So we have:
309 = abc + ab
If b + a = 0, a = 3 is impossible.
Therefore a = 2, b = 8, c = 1. So abc = 281.
This December I participated in Advent of Code.
This is a case of, part 1 can be brute forced, part 2 can’t. Part 1 is straightforward. Simply keep a list of the numbers from each step. In part 2 it matters, that the order doesn’t matter. If I may say so. Instead of keeping a list of the numbers, I keep a count. How many 0s, how many 1s etc. So e.g. in a step I say, I previously had x 0s, now I have x 1s. I previously had y 1s, now I have y 2024s. I previously had z 2024s, now I have z 20s and z 24s. Etc. Adding everything up, because there might be more than one source for the numbers in the next step.
I liked this one!
Part 1:
<?php
$input = file_get_contents('./d11input2.txt', true);
foreach(preg_split("/((\r?\n)|(\r\n?))/", $input) as $line) {
if(strlen($line)>2) {
$num1 = explode(" ", $line);
}
}
/*
If the stone is engraved with the number 0, it is replaced by a stone engraved with the number 1.
If the stone is engraved with a number that has an even number of digits, it is replaced by two stones. The left half of the digits are engraved on the new left stone, and the right half of the digits are engraved on the new right stone. (The new numbers don't keep extra leading zeroes: 1000 would become stones 10 and 0.)
If none of the other rules apply, the stone is replaced by a new stone; the old stone's number multiplied by 2024 is engraved on the new stone.
*/
for($i=0;$i<25;$i++) {
for($j=0;$j<sizeof($num1);$j++) {
$num = $num1[$j];
if($num == 0) {
$num2[] = 1;
continue;
}
$numl = strlen((string) $num);
if($numl % 2 == 0) {
$num2[] = (int) substr((string) $num, 0, $numl/2);
$num2[] = (int) substr((string) $num, $numl/2, $numl/2);
continue;
}
$num2[] = $num * 2024;
}
$num1 = $num2;
$num2 = array();
}
print(sizeof($num1)."\n");
?>Part 2:
<?php
$input = file_get_contents('./d11input2.txt', true);
foreach(preg_split("/((\r?\n)|(\r\n?))/", $input) as $line) {
if(strlen($line)>2) {
$num1 = explode(" ", $line);
}
}
for($j=0;$j<sizeof($num1);$j++) {
$num = $num1[$j];
if(isset($num2[$num])) {
$num2[$num]++;
} else {
$num2[$num] = 1;
}
}
$num1 = $num2;
$num2 = array();
/*
If the stone is engraved with the number 0, it is replaced by a stone engraved with the number 1.
If the stone is engraved with a number that has an even number of digits, it is replaced by two stones. The left half of the digits are engraved on the new left stone, and the right half of the digits are engraved on the new right stone. (The new numbers don't keep extra leading zeroes: 1000 would become stones 10 and 0.)
If none of the other rules apply, the stone is replaced by a new stone; the old stone's number multiplied by 2024 is engraved on the new stone.
*/
for($i=0;$i<75;$i++) {
foreach($num1 as $num => $nom) {
if($num == 0) {
if(isset($num2[1])) {
$num2[1] += $nom;
} else {
$num2[1] = $nom;
}
continue;
}
$numl = strlen((string) $num);
if($numl % 2 == 0) {
$numa = (int) substr((string) $num, 0, $numl/2);
if(isset($num2[$numa])) {
$num2[$numa] += $nom;
} else {
$num2[$numa] = $nom;
}
$numb = (int) substr((string) $num, $numl/2, $numl/2);
if(isset($num2[$numb])) {
$num2[$numb] += $nom;
} else {
$num2[$numb] = $nom;
}
continue;
}
$numc = $num * 2024;
if(isset($num2[$numc])) {
$num2[$numc] += $nom;
} else {
$num2[$numc] = $nom;
}
}
$num1 = $num2;
$num2 = array();
}
print(array_sum($num1)."\n");
?>A new December and a new bunch of puzzles from mscroggs.co.uk.
Out of the 11 available integers, 5 is the 5th, with 4 integers before it and 6 after. A set of integers is well formed, if 5 is the middle integer, with n integers before it and n after.
n = 0, 1 way to do that.
n = 1, 4 choices in front, 6 choices after, 4*6 = 24 ways to do that.
n = 2, (4 c 2) choices in front, (6 c 2) choices after, (4 c 2) * (6 c 2) = 4*3/2 * 6*5/2 = 6*15 = 90 ways to do that.
n = 3, (4 c 3) choices in front, (6 c 3) choices after, (4 c 3) * (6 * 3) = 4 * 6*5*4/(3*2) = 4*20 = 80 ways to do that.
n = 4, 1 choice in front, (6 c 4) choices after, 6*5/2 = 15 ways to do that.
1+24+90+80+15 = 210.
Stuff from ommadawn.dk (Lise Andreasen) 