Question

In a village, there are 87 families of which 52 families have at most 2 children. In a rural development program, 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made?

(a) \[{}^{52}{C_{18}} \times {}^{35}{C_2} + {}^{52}{C_{19}} \times {}^{35}{C_1} + {}^{52}{C_{20}} \times {}^{35}{C_0}\]

(b) \[{}^{52}{C_{18}} \times {}^{35}{C_2} + {}^{52}{C_{19}} \times {}^{35}{C_1} + {}^{52}{C_{20}} \times {}^{35}{C_{10}}\]

(c) \[{}^{52}{C_{18}} \times {}^{35}{C_2} + {}^{52}{C_{19}} \times {}^{35}{C_1} + {}^{52}{C_{20}} \times {}^5{C_0}\]

(d) \[{}^{52}{C_{18}} \times {}^{35}{C_2} + {}^{52}{C_9} \times {}^{35}{C_{10}} + {}^{52}{C_{20}} \times {}^5{C_0}\]

Answer 1

Hint: Divide the problem into three cases, choosing 18 families having at most 2 children plus 2 families not having at most 2 children, choosing 19 families having at most 2 children plus 1 family not having at most 2 children, and choosing 20 families having at most 2 children. Then use the combination to find the number of ways.

Complete step-by-step answer:
It is given that there are 87 families in total out of which 52 families have at most 2 children.
Then the number of families with more than 2 children is 87 – 52 which is 35.

For the rural development program, 20 families are to be chosen, of which at least 18 families must have at most 2 children.

This can be divided into three cases.

In case 1, 18 families with at most two children are chosen along with 2 families with more than 2 children.

\[{N_1} = {}^{52}{C_{18}} \times {}^{35}{C_2}................(1)\]

In case 2, 19 families with at most two children are chosen along with 1 family with more than 2 children.

\[{N_2} = {}^{52}{C_{19}} \times {}^{35}{C_1}.............(2)\]

In case 3, all 20 families are selected from families with at most 2 children and hence, the number of families with more than 2 children is zero.

\[{N_3} = {}^{52}{C_{20}} \times {}^{35}{C_0}.............(3)\]

The total number of ways is obtained by adding equations (1), (2), and (3).

\[N = {N_1} + {N_2} + {N_3}\]

\[N = {}^{52}{C_{18}} \times {}^{35}{C_2} + {}^{52}{C_{19}} \times {}^{35}{C_1} + {}^{52}{C_{20}} \times {}^{35}{C_0}\]

Hence, the correct answer is option (a).

Note: While multiplying the \[{}^n{C_r}\] terms, check that the sum of objects chosen is 20, for example in \[{}^{52}{C_{18}} \times {}^{35}{C_2}\], 18 + 2 is 20. You can also use this trick to select the answer directly.