We must show that P(k + 1) is true. Select sentences from the following scrambled list and put them in the correct order to continue the proof. By inductive hypothesis, the number of 1's in t is a multiple of 3 and, therefore, equals 3a for some integer a.Now t contains 3a 1's and each of 1110 and 0111 contains three 1's. So the total number of 1's in s is 3a + 3, which equals 3(a + 1) and is a multiple of 3.Since either 1110 or 0111 was removed from s to obtain t, then t is a string in S that has length 4k and consists of juxtapositions of 1110 and 0111.Suppose s is any string in S that has length 4(k + 1). We must show that the number of 1's in s is a multiple of 3.Since s has length 4(k + 1) = 4k + 4, either the first four or the final four characters in s are 1110 or 0111. Let t be the string that is the remaining part of s. Continue the proof below. By inductive hypothesis, the number of 1's in t is a multiple of 3 and, therefore, equals 3a for some integer a. Incorrect: Your answer is incorrect. All the strings in S are obtained by juxtaposing strings of the form 1110 and 0111. Now t contains 3a 1's and each of 1110 and 0111 contains three 1's. So the total number of 1's in s is 3a + 3, which equals 3(a + 1) and is a multiple of 3. Incorrect: Your answer is incorrect. ---Select--- Incorrect: Your answer is incorrect. ---Select--- Incorrect: Your answer is incorrect. ---Select--- Incorrect: Your answer is incorrect. Hence, P(k + 1) is true, and so the inductive step is complete.