Count how many times t appears as a subsequence of s. Let dp[i][j] = count for s[0..i−1] and t[0..j−1]. If s[i−1]=t[j−1]: we can match this character (dp[i−1][j−1]) or skip it (dp[i−1][j]). Total: dp[i][j]=dp[i−1][j−1]+dp[i−1][j]. If they don't match: must skip s[i−1]. dp[i][j]=dp[i−1][j]. Example: s="rabbbit", t="rabbit".
Answer: 3 (three ways to choose which 'b' to skip).