To solve this problem, we will use the concept of combinations in mathematics.

Step 1: Exclude the 2 oldest boys from the selection pool. This leaves us with 6 boys to choose from.

Step 2: We need to choose 4 boys out of these 6.

Step 3: The number of ways to choose 4 boys out of 6 (without repetition) is given by the combination formula: C(n, r) = n! / [(n-r)! * r!], where n is the total number of items, r is the number of items to choose, and '!' denotes factorial.

Step 4: Substituting the values into the formula, we get: C(6, 4) = 6! / [(6-4)! * 4!] = 15.

So, there are 15 different ways the committees can be arranged.

Question

To solve this problem, we will use the concept of combinations in mathematics.

Step 1: Exclude the 2 oldest boys from the selection pool. This leaves us with 6 boys to choose from.

Step 2: We need to choose 4 boys out of these 6.

Step 3: The number of ways to choose 4 boys out of 6 (without repetition) is given by the combination formula: C(n, r) = n! / [(n-r)! * r!], where n is the total number of items, r is the number of items to choose, and '!' denotes factorial.

Step 4: Substituting the values into the formula, we get: C(6, 4) = 6! / [(6-4)! * 4!] = 15.

So, there are 15 different ways the committees can be arranged.

Knowee AI · Accepted Answer

To solve this problem, we will use the concept of combinations in mathematics.

Step 1: Exclude the 2 oldest boys from the selection pool. This leaves us with 6 boys to choose from.

Step 2: We need to choose 4 boys out of these 6.

Step 3: The number of ways to choose 4 boys out of 6 (without repetition) is given by the combination formula: C(n, r) = n! / [(n-r)! * r!], where n is the total number of items, r is the number of items to choose, and '!' denotes factorial.

Step 4: Substituting the values into the formula, we get: C(6, 4) = 6! / [(6-4)! * 4!] = 15.

So, there are 15 different ways the committees can be arranged.

There are 8 boys. Only 4 can be chosen for a committee. 2 of the oldest boys cannot be chosen. There cannot be any repetitions of the boys in each committee. Find the number of ways the committees can be arranged

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