Question

(c) Out of 5 mathematicians and 7 physicists, a committee consisting of 2 mathemat
and 3 physicists is to be formed. In how many ways can this be done if
(i) any mathematician and any physicist can be included.
(ii) one particular physicist must be on the committee
(iii) two particular mathematicians cannot be on the committee
16. Five red marbles, two white marbles, and three blue marbles are arranged in a row. It
the marbles of the same colour are not distinguishable from each other, how many differet
arrangements are possible​

Answer 1

Answer:

Out of 5 Mathematicians and 7 physicists, a committee consisting of 2 mathematicians and 3 physicists is to be formed. In how many ways can this be done if. b. 5C2*6C2*1=150, because the specific physicist takes 1 from each part.

Answer 2

Concept Introduction:-

Combinations are a means of picking objects or numbers from a set of objects or collection and Permutation is the act of putting objects and numbers in order.

Given Information:-

$(c)$ We have been given that $5$ mathematicians and $7$ physicists, a committee consisting of $2$ mathematics and $3$ physicists is to be formed.

$16$  We have been given that Five red marbles, two white marbles, and three blue marbles are arranged in a row.

To Find:-

$(c)$ We need to find in how many ways can this be done if

$(i)$ any mathematician and any physicist can be included.

$(ii)$ one particular physicist must be on the committee

$(iii)$ two particular mathematicians cannot be on the committee

$16$ We need to find how many different arrangements are possible​

Solution:-

According to the problem

$(c)$ Total number of ways$=5_{C_{2}\times7_{C_{3}\\=10\times35\\=350$

$16$ All $10$ marbles can be arranged (permuted) in $10!$ ways. Many of these arrangements are identical because same colored marbles are indistinguishable from one another. In order to arrive at only those permutations that are distinct, division is required by each permutation relative to blue, green and white marbles, which are $5!$, $2!$ and $3!$, respectively. Thus the number of distinct permutations of these $10$ marbles $= 10!/(5!)(2!)(3!) = 2,520$

Final Answer:-

The correct answer are

$(c)$ Total number of ways is $350$.

$16$) $2520$ different arrangements are possible​.

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