MHTCET 15 Apr 2026 Shift 2 Paper

Concept:The 7 identical balls have no individual identities, so only their count in the selection matters.Explanation:Out of 23 balls, $23-7=16$ balls are distinct.We need to select 12 balls in total.Let $r$ identical balls be selected, where $r=0,1,2,\ldots,7$.Then $12-r$ balls must be selected from the 16 distinct balls.Hence the total number of ways is$$\sum_{r=0}^{7}\binom{16}{12-r}=\sum_{k=5}^{12}\binom{16}{k}.$$Writing the sum explicitly and using $\binom{n}{r}=\binom{n}{n-r}$, we get$$\binom{16}{4}+2\binom{16}{5}+2\binom{16}{6}+2\binom{16}{7}+\binom{16}{8}.$$By Pascal's identity,$$\binom{18}{6}=\binom{16}{4}+2\binom{16}{5}+\binom{16}{6}$$and$$\binom{18}{8}=\binom{16}{6}+2\binom{16}{7}+\binom{16}{8}.$$Adding these gives the same expression.Therefore, the required number of selections is$$\binom{18}{6}+\binom{18}{8}.$$Answer:Option A: ${ }^{18}C_6+{ }^{18}C_8$.