從arr中挑選出來需要保存的(arr.length - count)所有可能的組合，然后再從原數組中挑選出來所有可能的(count)補充組合，然后拼在一起去重就行了。

(async ()=>{

let vari = ['A','T','C','G'];

function change(arr, count = 1){

arr = arr.map(a=>vari.indexOf(a)).sort(); // convert to int array, for easier sorting in future

var prefix = combination(arr, arr.length - count);

var suffix = combination([0,1,2,3], count);

var res = prefix.length === 0 ? suffix : prefix.map(p=>suffix.map(s=>[...p, ...s])).reduce((a,b)=>[...a,...b], []);

var existing = arr.join(",");

res = res

.filter(r=>new Set(r).size === r.length) // filter out ones with duplicated items, like ["A","A","C"]

.map(r=>r.sort()) // order items, so ["C","A"] becomes ["A", "c"]

.sort() // sort, just make it looks prettier

.reduce((a,b)=>{ // remove duplicated combinations, keep only one

if(a[b.join(",")] === void 0){

a.arr.push(b);

a[b.join(",")] = 0;

}

return a;

}, {arr: []}).arr

.filter(r=>r.join(",") !== existing);

return res.map(r=>r.map((i)=>vari[i]));// convert back to letters

}

function combination(src, pick){

if(pick > src.length){

return [];

}

if(pick === 1){

return src.map(s=>[s]);

}

var res = [];

for(var i=0;i

for(var r of combination(src.slice(i+1), pick - 1)){

res.push([src[i], ...r]);

}

}

return res;

}

console.log(change(["G","A","C"], 2));

})();

output [A,T,C],[A,T,G],[T,C,G]

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