gulp combining tasks into default command

How to I go about combining my styles task my browser-sync and my watch task into one gulp default command?

Here is what I have so far.

var gulp = require("gulp");
var sass = require("gulp-sass");
var browserSync = require("browser-sync").create();

gulp.task("styles", function() {
  gulp
    .src("sass/**/*.scss")
    .pipe(sass().on("error", sass.logError))
    .pipe(gulp.dest("./css/"));
});

// Static server
gulp.task("browser-sync", function() {
  browserSync.init({
    server: {
      baseDir: "./"
    }
  });
});

//Watch task
gulp.task("default", function() {
  gulp.watch("sass/**/*.scss", gulp.series("styles"));
});

I have also tried this from the browser-sync documentation but it gives me the error AssertionError [ERR_ASSERTION]: Task function must be specified With the same variables at the top.

gulp.task("serve", ["sass"], function() {
  browserSync.init({
    server: "./"
  });

  gulp.watch("/sass/*.scss", ["sass"]);
  gulp.watch("/*.html").on("change", browserSync.reload);
});

// Compile sass into CSS & auto-inject into browsers
gulp.task("sass", function() {
  return gulp
    .src("/scss/*.scss")
    .pipe(sass())
    .pipe(gulp.dest("/css"))
    .pipe(browserSync.stream());
});

gulp.task("default", ["serve"]);

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