Question

Tanuj has 9 relatives, 5 of them are Ladies and 4 Gentlemen; his wife has also 9 relatives, 4 of them are Ladies and 5 Gentlemen and there are no common relatives for Tanuj and his wife. If the number of ways they can invite a dinner party of 4 Ladies and 4 Gentlemen so that there are 4 of the Tanuj relatives and 4 of his wife's relatives is m then sum of digits of m equals to

• A. 19
• B. 18
• C. 20
• D. 17

Answer 1

Answer: (A)

19

Answer 2

Let $T_L=5$ and $T_G=4$ be the number of ladies and gentlemen among Tanuj's relatives, respectively. Let $W_L=4$ and $W_G=5$ be the number of ladies and gentlemen among his wife's relatives, respectively.

The party needs to have 4 ladies and 4 gentlemen. Also, 4 guests must be from Tanuj's relatives and 4 from his wife's relatives.

Let $t_l$ and $t_g$ be the number of ladies and gentlemen invited from Tanuj's relatives. So, $t_l + t_g = 4$.

Let $w_l$ and $w_g$ be the number of ladies and gentlemen invited from his wife's relatives. So, $w_l + w_g = 4$.

The total number of ladies in the party is 4: $t_l + w_l = 4$. The total number of gentlemen in the party is 4: $t_g + w_g = 4$.

From these equations, we get $w_l = 4 - t_l$ and $w_g = 4 - t_g$. Also, substituting $t_l = 4 - t_g$ into $t_l + w_l = 4$, we get $(4 - t_g) + w_l = 4$, which implies $w_l = t_g$. Similarly, substituting $t_g = 4 - t_l$ into $t_g + w_g = 4$, we get $(4 - t_l) + w_g = 4$, which implies $w_g = t_l$.

We need to find the number of ways to choose $t_l$ ladies from Tanuj's 5 ladies and $t_g$ gentlemen from Tanuj's 4 gentlemen, such that $t_l + t_g = 4$. The number of ways for Tanuj's selection is ${}^5C_{t_l} \times {}^4C_{t_g}$. Simultaneously, we need to choose $w_l = t_g$ ladies from his wife's 4 ladies and $w_g = t_l$ gentlemen from his wife's 5 gentlemen. The number of ways for his wife's selection is ${}^4C_{w_l} \times {}^5C_{w_g} = {}^4C_{t_g} \times {}^5C_{t_l}$.

The total number of ways for a specific pair $(t_l, t_g)$ is $({}^5C_{t_l} \times {}^4C_{t_g}) \times ({}^4C_{t_g} \times {}^5C_{t_l})$.

We consider all valid combinations of $(t_l, t_g)$ such that $t_l + t_g = 4$, and the selections are possible: $0 \le t_l \le 5$, $0 \le t_g \le 4$, $0 \le w_l \le 4$, $0 \le w_g \le 5$. These constraints imply $0 \le t_l \le 4$.

Case 1: $t_l=0, t_g=4$. Ways = $({}^5C_0 \times {}^4C_4) \times ({}^4C_4 \times {}^5C_0) = (1 \times 1) \times (1 \times 1) = 1$.

Case 2: $t_l=1, t_g=3$. Ways = $({}^5C_1 \times {}^4C_3) \times ({}^4C_3 \times {}^5C_1) = (5 \times 4) \times (4 \times 5) = 20 \times 20 = 400$.

Case 3: $t_l=2, t_g=2$. Ways = $({}^5C_2 \times {}^4C_2) \times ({}^4C_2 \times {}^5C_2) = (10 \times 6) \times (6 \times 10) = 60 \times 60 = 3600$.

Case 4: $t_l=3, t_g=1$. Ways = $({}^5C_3 \times {}^4C_1) \times ({}^4C_1 \times {}^5C_3) = (10 \times 4) \times (4 \times 10) = 40 \times 40 = 1600$.

Case 5: $t_l=4, t_g=0$. Ways = $({}^5C_4 \times {}^4C_0) \times ({}^4C_0 \times {}^5C_4) = (5 \times 1) \times (1 \times 5) = 5 \times 5 = 25$.

The total number of ways, $m$, is the sum of ways from all cases: $m = 1 + 400 + 3600 + 1600 + 25 = 5626$.

The sum of the digits of $m$ is $5 + 6 + 2 + 6 = 19$.