The correct option is B 1
=∣∣
∣
∣∣111nC1n+1C1n+2C1nC2n+1C2n+2C2∣∣
∣
∣∣
=∣∣
∣
∣∣111nC1n+1C1n+1C0+n+1C1nC2n+1C2n+1C1+n+1C2∣∣
∣
∣∣
C3→C3−C2
=∣∣
∣
∣∣110nC1n+1C1n+1C0nC2n+1C2n+1C1∣∣
∣
∣∣
Similarly,
C2→C2−C1
=∣∣
∣
∣∣100nC1nC0n+1C0nC2nC1n+1C1∣∣
∣
∣∣
=nC0×n+1C1−n+1C0×nC1=n+1−n=1