A-delegation-of-4-people-is-to-be-selected-from-5-women-and-6-men-Find-the-number-of-possible-delegations-if-a-there-are-no-restrictions-b-there-is-at-least-1-woman-c-there-are-at-least-2-w

Question Number 5280 by Rasheed Soomro last updated on 04/May/16

A delegation of 4 people is to be

selected from 5 women and 6 men .

Find the number of possible delegations if

(a) there are no restrictions,

(b) there is at least 1 woman,

(c) there are at least 2 women .

One of the men cannot get along with

one of the women . Find the number of

delegations which include this particular

man or woman, but not both .

Answered by Yozzii last updated on 04/May/16

(a) no . of ways = (\begin{matrix} 5 + 6 \\ 4 \end{matrix}) = (\begin{matrix} 11 \\ 4 \end{matrix}) = 330

(b) no . of ways = total no . of possible combinations - no . of groups with only men

= (\begin{matrix} 5 + 6 \\ 4 \end{matrix}) - (\begin{matrix} 6 \\ 4 \end{matrix})

= 315

Set theoretically, if we know the cardinality

of the finite universal set U and the cardinality

of one of two disjoint sets A and B, and

U = A \cup B, then ∣ A ∣ + ∣ B ∣=∣ U ∣ .

This situation has

U = all possible delegations

A = delegations with only men,

B = delegations with at least one woman .

\Rightarrow∣ B ∣=∣ U ∣ - ∣ A ∣ .

(c) W e can subtract from the total number

of possible combinations those combinations

that are all men and those combinations

that include 3 men and 1 woman .

By the And counting principle,

no . of combinations of 3 men and 1 woman

is equal to (\begin{matrix} 6 \\ 3 \end{matrix}) \times (\begin{matrix} 5 \\ 1 \end{matrix}) .

\Rightarrow no . of ways required = (\begin{matrix} 5 + 6 \\ 4 \end{matrix}) - {(\begin{matrix} 6 \\ 4 \end{matrix}) + (\begin{matrix} 6 \\ 3 \end{matrix}) \times (\begin{matrix} 5 \\ 1 \end{matrix})}

= 315 - 100

= 215

(d) no . of ways = (\begin{matrix} 11 - 2 \\ 3 \end{matrix}) + (\begin{matrix} 11 - 2 \\ 3 \end{matrix}) = 2 (\begin{matrix} 9 \\ 3 \end{matrix}) = 168

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