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(A) group of seven persons is to be formed from $8$ men and $6$ women. The condition is that there must be at least two women in the group.The possibl

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$3256$

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(A) group of seven persons is to be formed from $8$ men and $6$ women. The condition is that there must be at least two women in the group.The possible combinations are $(2W, 5M), (3W, 4M), (4W, 3M), (5W, 2M),$ and $(6W, 1M)$.The total number of ways is given by the sum of these combinations:$= {}^{6}C_{2} 	imes {}^{8}C_{5} + {}^{6}C_{3} 	imes {}^{8}C_{4} + {}^{6}C_{4} 	imes {}^{8}C_{3} + {}^{6}C_{5} 	imes {}^{8}C_{2} + {}^{6}C_{6} 	imes {}^{8}C_{1}$Calculating each term:$= (15 	imes 56) + (20 	imes 70) + (15 	imes 56) + (6 	imes 28) + (1 	imes 8)$$= 840 + 1400 + 840 + 168 + 8 = 3256$Alternatively,total ways to choose $7$ from $14$ is ${}^{14}C_{7} = 3432$. Subtract cases with $0$ women $({}^{6}C_{0} 	imes {}^{8}C_{7} = 1 	imes 8 = 8)$ and $1$ woman $({}^{6}C_{1} 	imes {}^{8}C_{6} = 6 	imes 28 = 168)$.$3432 - (8 + 168) = 3432 - 176 = 3256$.Therefore,the required number of groups is $3256$.

From $8$ men and $6$ women,a group of seven persons is to be formed. In how many ways can the group be formed such that the group includes at least two ladies?

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