A planning committee for a major development project must consist six members. There are eight architects and five engineers available to choose from. In how many different ways can this planning committee be formed (rounded off to zero decimals)?

A planning committee for a major development project must consist six members. There are eight architects and five engineers available to choose from. In how many different ways can this planning committee be formed (rounded off to zero decimals)?

Question
Decimals
asked 2021-01-23
A planning committee for a major development project must consist six members. There are eight architects and five engineers available to choose from. In how many different ways can this planning committee be formed (rounded off to zero decimals)?

Answers (1)

2021-01-24
Step 1
Introduction:
Combination:
Suppose there are n distinct items in a population. From these items, r distinct items are to be chosen, without any added constraints, in such a way that the order of choosing does not have any particular importance.
Then, the number of possible ways in which, the r distinct items can be chosen from the n distinct items in the population is “n combination r”, denoted as: \(\displaystyle_{n}{C}_{{r}}={n}!\text{/}{\left[{r}!{\left({n}-{r}\right)}!\right]}.\)
Step 2
Calculation:
There are 8 architects and 5 engineers. Thus, total number of individuals in the population is,
\(\displaystyle{n}={13}{\left(={8}+{5}\right)}.\)
Out of these individuals, \(r = 6\) members are to be chosen for a planning committee. Evidently, 6 distinct individuals must be chosen. There are no added constraints.
The number of ways in which the 6 members can be chosen from the 13 architects and engineers is:
\(\displaystyle_{13}{C}_{{6}}\)
\(\displaystyle={13}!\text{/}{\left[{6}!{\left({13}-{6}\right)}!\right]}\)
\(\displaystyle={13}!\text{/}{\left[{\left({6}!\right)}{\left({7}!\right)}\right]}\)
\(=1.716\)
Thus, the number of ways in which the 6 member-committee can be formed is 1.716
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