A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then :

Permutations and Combinations PYQ — JEE Main 2019 · 9 Apr · Shift 1 · Q43 with Solution

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Correct answer: B

Total people available

There are $8$ males and $5$ females, so total people $= 13$ . A committee of $11$ members is to be formed.

Find $m$ : committees with at least $6$ males

Let the number of males in the committee be $k$ . Since the committee has $11$ members, number of females will be $11-k$ .

Condition: at least 6 males, so $k \ge 6$ . Also, there are only $5$ females available, so the committee must leave out exactly $2$ people total from the $13$ .

Possible male-female compositions for an 11-member committee are:

$6M, 5F$
$7M, 4F$
$8M, 3F$

Now count each case:

\binom{8}{6}\binom{5}{5} + \binom{8}{7}\binom{5}{4} + \binom{8}{8}\binom{5}{3}

Compute:

\binom{8}{6} = 28, \quad \binom{5}{5}=1

\binom{8}{7}=8, \quad \binom{5}{4}=5

\binom{8}{8}=1, \quad \binom{5}{3}=10

So,

m = 28\cdot 1 + 8\cdot 5 + 1\cdot 10 = 28 + 40 + 10 = 78

Find $n$ : committees with at least $3$ females

Let the number of females be $r$ . Condition: at least 3 females, so $r \ge 3$ .

Possible compositions are:

$8M, 3F$
$7M, 4F$
$6M, 5F$

Thus,

n = \binom{8}{8}\binom{5}{3} + \binom{8}{7}\binom{5}{4} + \binom{8}{6}\binom{5}{5}

Compute:

n = 1\cdot 10 + 8\cdot 5 + 28\cdot 1 = 10 + 40 + 28 = 78

So,

n = 78

Compare with options

We found:

m = 78, \quad n = 78

Check options:

A: $n = m - 8 \Rightarrow 78 = 70$ ❌
B: $m = n = 78$ ✅
C: $m+n=68 \Rightarrow 156=68$ ❌
D: $m=n=68$ ❌

Final answer

The correct option is B.

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View written solutionFree

Correct answer: B

Total people available

There are $8$ males and $5$ females, so total people $= 13$ . A committee of $11$ members is to be formed.

Find $m$ : committees with at least $6$ males

Let the number of males in the committee be $k$ . Since the committee has $11$ members, number of females will be $11-k$ .

Condition: at least 6 males, so $k \ge 6$ . Also, there are only $5$ females available, so the committee must leave out exactly $2$ people total from the $13$ .

Possible male-female compositions for an 11-member committee are:

$6M, 5F$
$7M, 4F$
$8M, 3F$

Now count each case:

\binom{8}{6}\binom{5}{5} + \binom{8}{7}\binom{5}{4} + \binom{8}{8}\binom{5}{3}

Compute:

\binom{8}{6} = 28, \quad \binom{5}{5}=1

\binom{8}{7}=8, \quad \binom{5}{4}=5

\binom{8}{8}=1, \quad \binom{5}{3}=10

So,

m = 28\cdot 1 + 8\cdot 5 + 1\cdot 10 = 28 + 40 + 10 = 78

Find $n$ : committees with at least $3$ females

Let the number of females be $r$ . Condition: at least 3 females, so $r \ge 3$ .

Possible compositions are:

$8M, 3F$
$7M, 4F$
$6M, 5F$

Thus,

n = \binom{8}{8}\binom{5}{3} + \binom{8}{7}\binom{5}{4} + \binom{8}{6}\binom{5}{5}

Compute:

n = 1\cdot 10 + 8\cdot 5 + 28\cdot 1 = 10 + 40 + 28 = 78

So,

n = 78

Compare with options

We found:

m = 78, \quad n = 78

Check options:

A: $n = m - 8 \Rightarrow 78 = 70$ ❌
B: $m = n = 78$ ✅
C: $m+n=68 \Rightarrow 156=68$ ❌
D: $m=n=68$ ❌

Final answer

The correct option is B.

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