(a+b+c+d)^{2} | |
a^{2}+b^{2}+c^{2}+d^{2} | With 2 factors, n=2, we can choose two "a" in one way: one from each factor, so the coefficient of the a^{2} term is 1. Similarly for the rest of the squared terms. Alternatively, we have C(2,2)=1. |
+2ab | For the terms containing ab, we can choose an "a" in 2 ways (one from the first factor, or one from the second factor). We can then choose a "b" in one way from the remaining factor, giving the coefficient of ab as 2·1=2. Alternatively, C(2,1)·C1,1)=2. |
+2ac+2ad | Similarly, we have the other terms containing a and another letter: |
+2bc+2bc+2cd | Similarly for those containing b, c and d |
Check the coefficients add up to (1+1+1+1)^{2}= 16 |
(a+b+c+d)^{2} | |
a^{2}+b^{2}+c^{2}+d^{2} | Write down the squares |
2ab+2ac+2ac | Then, starting with a, write down twice a times the other letters: 2a(b+c+d) |
2bc+2bd | Similarly for b, with the remaining letters: 2b(c+d) |
2cd | Finally with c. ("Finally" because no letters follow d). 2c(d) |
Check the sum of the coefficients is equal to (1+1+1+1)^{2}=16. |
(a+b+c)^{3} | |
a^{3}+b^{3}+c^{3} | We can choose three "a"'s for the cube in one way C(3,3)=1, or we can choose an a from the first factor and one from the second and one from the third, being the only way to make a3. The coefficient of the cubes is therefore 1. (It's the same for a, b and c, of course) |
3a^{2}b+3a^{2}c | Next, we consider the a^{2} terms. We can choose two a's from 3 factors in C(3,2) ways=3. We can choose a remaining letter in 1 way, so the coefficients of a^{2} are 3·1 ways. 3a^{2}(b+c) |
3ab^{2}+3b^{2}c | Similarly for the b^{2} terms: 3b^{2}(a+c) |
3ac^{2}+3bc^{2} | And the c^{2}: 3c^{2}(a+b) |
6abc | The remaining terms are abc's. We can choose an a in 3 ways, and then a b in 2 ways, and then we have only one way to choose a c. The coefficient is therefore 3·2·1=6 |
Finally, add up the coefficients to check they come to (1+1+1)^{3}=27 |
(a+b+c+d)^{3} | ||
a^{3}+b^{3}+c^{3}+d^{3} | Considering the cubes of a, we can choose 3 a's from 3 factors in one way only. C(3,3)=1. Similarly for the other letters | From 4 elements, we can choose 1, C(4,1)=4, so there are four terms here. The sum of the coefficients is 4. |
3a^{2}b+3a²c+3a²c | Considering a^{2}, we can choose 2 a's from 3 factors in C(3,2)=3 ways. We can write down the terms containing squares: |
As we can choose 1 element from 4, there are 4 possibilities here. (One shown to the left and the other three below and to the left) |
3b²a+3b²c+3b²c+ 3c²a+3c²b+3c²d+ 3d²a+3d²b+3d²b |
Similarly for the other letters. | For the third part, we can choose this letter from the remaining 3, C(3,1)=3. We therefore have 4 of 3 with a coefficient of 3, giving a sum of the coefficients of 36. |
6abc+6abd+6acd+6bcd | Finally, there are the combination of single letters, such as abc. We can choose the first letter from 3 factors in 3 ways. The second in 2. And the last in 1 way. That is C(3,1)·C(2,1)·C(1,1)=6. We therefore write the combinations down. | There are four elements, a, b, c, d so we can choose 3 of them from 4, in 4 ways. There are therefore four of these. The sum of the coefficients is 4·6=24. |
We add up the coefficients to check that they sum to (1+1+1+1)^{3}=64 | We can also check the part sums, as above |
(a+b+c+d+e)^{3} | ||
a^{3}+b^{3}+c^{3}+d^{3}+e^{3} | As usual, the cubics have a coefficient 1: | There are 5 different letters, so we have 5 cubics. The sum of the coefficients is 1·5=5. |
3a²b+3a²c+3a²d+3a²e+ 3b²a+3b²c+3b²d+3b²e+ 3c²a+3c²b+3c²d+3a²e+ 3d²a+3d²b+3d²c+3d²e+ 3e²a+3e²b+3e²c+3e²d+ |
Considering the a^{2} terms, we can choose an a^{2} term by selecting 2 a's from the three factors, C(3,2)=3. | We can choose a letter for a square in 5 ways, which we can combine with 4 other letters, to make 5·4=20. The sum of the coefficients is: 20·3=60 |
6abc+6abd+6abe+ 6acd+6ace+6ade+ 6bcd+6bce+ 6bde+ 6cde |
We can choose one of the letters in 3 ways; the second in 2; and the last in 1 way, making 6: C(3,1)·C(2,1)·C(1,1)=6 | We can select 3 letters from 5 in C(5,3) ways, or 10. The sum of the coefficients is therefore: 6·10=60 |
Check the sum of the coefficients. (1+1+1+1+1)^{3}=125 | We can also check the part sums as above. |
(a+b+c+d+e)^{4} | For comments, x∈{a,b,c,d,e} | |
a^{4}+b^{4}+c^{4}+d^{4}+e^{4}+ | We can choose 4 letters from 4, in one way, so the coefficient of x^{4} is C(4,4)=1. | We can choose 1 letter from 5 elements in 5 ways. |
4a^{3}(b+c+d+e)+ 4b^{3}(a+c+d+e)+ 4c^{3}(a+b+d+e)+ 4d^{3}(a+b+c+e)+ 4e^{3}(a+b+c+d)+ |
x^{3}: We can choose three x's from 4 factors in C(4,3) ways=4 | We can choose a letter (for the cube) from 5 elements in C(5,1) ways, and the remaining letter in C(4,1) ways. So there are 5⋅4=20 groups, and the sum of the coefficients is 20⋅4=80 |
6a^{2}(b^{2}+c^{2}+d^{2}+e^{2})+ 6b^{2}(c^{2}+d^{2}+e^{2})+ 6c^{2}(d^{2}+e^{2})+ 6d^{2}e^{2}+ |
We can choose x^{2} from 4 factors in C(4,2) ways. And then choose another in C(2,2) ways, giving 6⋅1 ways | We can choose 2 letters from 5 in C(5,2)=10 ways. This gives us 6⋅10=60 is the sum of the coefficients |
12a^{2}(bc+bd+be+cd+ce+de)+ 12b^{2}(ac+ad+ae+cd+ce+de)+ 12c^{2}(ab+ad+ae+bd+be+de)+ 12d^{2}(ab+ac+ae+bc+be+ce)+ 12e^{2}(ab+ac+ad+bc+bd+cd)+ |
We can choose x^{2} as above in 6 ways. From the remaining 2 factors, we can choose two different letters in C(2,1)⋅C(1,1) ways, giving in all 12 ways. | We can choose 1 letter from 5 in 5 ways. And 2 letters from 4 in C(4,2) ways giving 30 ways. The sum of the coefficients is therefore: 30⋅12=360 |
24abcd+24abce+24abde+24acde+ 24bcde |
For letters we can choose C(4,1)⋅C(3,1)⋅C(2,1)⋅C(1,1)=24 | We can choose four letters from 5 in C(5,4) ways=5. The coefficient sum is therefore 5⋅24=120 |
We check the sum of the coeffients is (1+1+1+1+1)^{4}=625 |