A group of 9 students,s_1, s_2, ldots, s_9,is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Total ways to form teams $X, Y, Z$ of sizes $2, 3, 4$ without any restrictions is $\binom{9}{2} 	imes \binom{7}{3} 	imes \binom{4}{4} = 36 	ime

Vedclass · Accepted Answer

$665$

Vedclass · Answer

(D) Total ways to form teams $X, Y, Z$ of sizes $2, 3, 4$ without any restrictions is $\binom{9}{2} 	imes \binom{7}{3} 	imes \binom{4}{4} = 36 	imes 35 = 1260$.Let $A$ be the set of ways where $s_1 \in X$ and $B$ be the set of ways where $s_2 \in Y$.We want to find the total ways minus the ways where ($s_1 \in X$ $OR$ $s_2 \in Y$).By the Principle of Inclusion-Exclusion: $|A \cup B| = |A| + |B| - |A \cap B|$.$|A|$ (ways where $s_1 \in X$): $s_1$ is fixed in $X$,we choose $1$ more from $8$ for $X$,then $3$ from $7$ for $Y$: $\binom{8}{1} 	imes \binom{7}{3} = 8 	imes 35 = 280$.$|B|$ (ways where $s_2 \in Y$): $s_2$ is fixed in $Y$,we choose $2$ from $8$ for $X$,then $2$ from $6$ for $Y$: $\binom{8}{2} 	imes \binom{6}{2} = 28 	imes 15 = 420$.$|A \cap B|$ (ways where $s_1 \in X$ $AND$ $s_2 \in Y$): $s_1$ fixed in $X$,$s_2$ fixed in $Y$,we choose $1$ from $7$ for $X$,then $2$ from $6$ for $Y$: $\binom{7}{1} 	imes \binom{6}{2} = 7 	imes 15 = 105$.$|A \cup B| = 280 + 420 - 105 = 595$.Total valid ways = $1260 - 595 = 665$.

$A$ group of $9$ students,$s_1, s_2, \ldots, s_9$,is to be divided into three teams $X, Y$,and $Z$ of sizes $2, 3$,and $4$,respectively. Suppose that $s_1$ cannot be selected for team $X$,and $s_2$ cannot be selected for team $Y$. The number of ways to form such teams is:

Similar Questions

The number of ways in which $15$ identical gold coins can be distributed among $3$ persons such that each one gets at least $3$ gold coins is

From a class of $25$ students,$10$ are to be chosen for an excursion party. There are $3$ students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Let $S = \{1, 2, 3, \ldots, 9\}$. For $k = 1, 2, \ldots, 5$,let $N_k$ be the number of subsets of $S$,each containing five elements out of which exactly $k$ are odd. Then $N_1 + N_2 + N_3 + N_4 + N_5 =$

Five students are selected from $n$ students such that the ratio of the number of ways in which $2$ particular students are selected to the number of ways $2$ particular students are not selected is $2:3$. Then the value of $n$ is

In a football championship,there were played $153$ matches. Every team played one match with each other. The number of teams participating in the championship is

Vedclass Test Series

Exam Paper Generator

Online Exam Module