Question

There are 6 boys and 6 girls. The number of ways to choose atleast 2 persons so that the combination include at least a boy and a girl is

• A. $2^{12}-2^7-35$
• B. 36
• C. $2^{12}-13$
• D. $2^{12}-2^7+1$

Answer 1

Answer: (D)

Number of ways = Total ways - no boy - no girl $+1$
$=\left({}^6{C}_0+{}^6{C}_1+\dots \dots +{}^6{C}_6\right)\times \left({}^6{C}_0+{}^6{C}_1+\dots \dots {}^6{C}_6\right)$
${}^6{C}_0\times \left({}^6{C}_0+{}^6{C}_1+{}^6{C}_2+\dots \dots \cdot {}^6{C}_6\right)-{}^6{C}_0\times \left({}^6{C}_0+\dots \dots +{}^6{C}_0\right)+1$
$=2^{12}-2^6-2^6+1=2^{12}-2^7+1$
Aliter:
Required number of ways $=\left(2^6-1\right)\left(2^6-1\right)=2^{12}-2^7+1$